Solve the afterward problems.

The ambit of a apple is abstinent to be 24 cm, with a accessible absurdity of 0.25 cm. Use the cogwheel (dV) to appraisal the best absurdity in the affected aggregate (Vtext{.})

Observe that (ds dr=frac{1}{8pi}text{.})

(ds dV=frac{72}{pi^2}text{.})

The breadth (A) of a aboveboard of ancillary breadth (s) is (A=s^2text{.}) Suppose (s) increases by an bulk (Delta s=dstext{.})

Draw a aboveboard and again allegorize the abundance (dA) on your diagram.

If (dA) is acclimated to almost (Delta Atext{,}) allegorize the absurdity of approximation on the aforementioned diagram.

Observe that (Delta A=A(s Delta s)-A(s)=2sDelta s (Delta s)^2) and (dA=2sDelta stext{.})

Let (f(x)=sqrt{(x 4)^3}text{.})

Find the beeline approximation to the action (f(x)=sqrt{(x 4)^3}) at (a=0text{.})

Use this approximation to appraisal the cardinal (sqrt{(3.95)^3}text{.}) Is your appraisal an aggrandize or an underestimate?

What is the concavity of the action (f(x)text{?})

(L(x)=8 3xtext{.})

(sqrt{(3.95)^3} approx 7.85text{.}) Underestimate.

Note that (f(0)=8text{.}) From (ds f'(x)=frac{3}{2}sqrt{x 4}) it folows that (f'(0)=3text{.}) Thus the linearization of (f) at (a=0) is (L(x)=8 3xtext{.})

For (x) “close” to (0) we accept that (ds f(x)=frac{3}{2}sqrt{(x 4)^3}approx L(x)text{.}) Thus (sqrt{(3.95)^3}=f(-0.05)approx L(-0.05)=8-0.15=7.85text{.}) Back (ds f”(x)=frac{3}{4sqrt{x 4}}>0) we achieve that, in the neighbourhood of (x=0text{,}) the blueprint of the action (f) is aloft the departure band at (x=0text{.}) Thus (L(-0.05)) is an underestimate.

Use beeline approximation to appraisal (sqrt[3]{65}text{.})

Use concavity to accompaniment if your appraisal in (a) is greater than or beneath than the exact amount of (sqrt[3]{65}text{.}) Explain.

(ds sqrt[3]{65}approx L(65)=frac{193}{48}text{.})

Overestimate.

Use beeline approximation to appraisal the amount of (sqrt[3]{26^2}text{.}) Express your acknowledgment a distinct atom (for example, (ds frac{16}{729})).

(ds sqrt[3]{26^2} approx frac{79}{9}text{.})

Let (ds f(x)=x^{frac{2}{3}}text{.}) Again (ds f(x)=frac{2}{3}x^{-frac{1}{3}}text{,}) (f(27)=9text{,}) and (ds f'(27)=frac{2}{9}text{.}) Hence the linearization of the action (f) at (a=27) is (ds L(x)=9 frac{2}{9}(x-27)text{.}) It follows that (ds sqrt[3]{26^2}=f(26)approx L(26)=9-frac{2}{9}=frac{79}{9}text{.}) (Note: MAPLE gives (ds frac{79}{9}approx 8.777777778) and (sqrt[3]{26^2}approx 8.776382955text{.}))

Use the beeline approximation to almost ((63)^{2/3}text{.}) Again use differentials to appraisal the error.

(ds (63)^{2/3}approxfrac{95}{6}text{.})

Let (ds f(x)=x^{frac{2}{3}}text{.}) Again (ds f(x)=frac{2}{3}x^{-frac{1}{3}}text{,}) (f(64)=16text{,}) and (ds f'(64)=frac{1}{6}text{.}) Hence the linearization of the action (f) at (a=64) is (ds L(x)=16 frac{1}{6}(x-64)text{.}) It follows that (ds (63)^{2/3} =f(63)approx L(63)=16-frac{1}{6}=frac{95}{6}text{.}) The absurdity is abutting to the complete amount of the cogwheel (ds |dy|=|f'(64)Delta x| = frac{1}{6}text{.}) (Note: MAPLE gives (ds frac{95}{6}approx 15.83333333) and (sqrt[3]{63^2}approx 15.83289626text{.}))

Use beeline approximation to appraisal the amount of (sqrt{80}text{.})

(ds sqrt{80}approxfrac{161}{18}text{.})

Let (ds f(x)=sqrt{x}text{.}) Again (ds f(x)=frac{1}{2sqrt{x}}text{,}) (f(81)=9text{,}) and (ds f'(81)=frac{1}{18}text{.}) Hence the linearization of the action (f) at (a=81) is (ds L(x)=9 frac{1}{18}(x-81)text{.}) It follows that (ds sqrt{80}=f(80)approx L(80)=9-frac{1}{18}=frac{161}{18}text{.}) (Note: MAPLE gives (ds frac{161}{18}approx 8.944444444) and (sqrt{80}approx 8.944271910text{.}))

Assume that (f) is action such that (f(5)=2) and (f^prime(5)=4text{.}) Application a beeline approximation to (f) abreast (x=5text{,}) acquisition an approximation to (f(4.9)text{.})

The linearization of the action (f) at (a=5) is (L(x)=2 4(x-5)text{.}) Thus (f(4.9)approx L(4.9)=2-0.4=1.6text{.})

Suppose that we don’t accept a blueprint for (g(x)) but we apperceive that (g(2)=-4) and (g^prime (x)=sqrt{x^2 5}) for all (xtext{.})

Use beeline approximation to appraisal (g(2.05)text{.})

Is your appraisal in allotment (a) beyond or abate than the absolute value? Explain.

(-3.85text{.})

Larger.

The linearization of the action (g) at (a=2) is (L(x)=-4 3(x-2)text{.}) Thus (g(2.05)approx L(2.05)=-3.85text{.})

From (ds g”(2)=frac{2}{3}>0) we achieve that the action (g) is biconcave bottomward at (a=2text{,}) i.e. the blueprint of the action lies beneath the departure line. Thus, the appraisal is beyond than the absolute value.

Let (f(x)=sqrt{1-x}text{.})

Find a beeline approximation for the action (f(x)=sqrt{1-x}) accurate for (x) abutting to (0text{.})

Use your acknowledgment to acquisition an almost amount for (sqrt{0.9}text{.})

Find the departure band to the blueprint of (f(x)=sqrt{1-x}) at (x=0text{.})

Sketch a blueprint to allegorize the accord amid (f(x)=sqrt{1-x}) and its beeline approximation abreast (x=0text{.})

(ds L(x)=1-frac{x}{2}text{.})

(ds sqrt{0.9}approx 1-frac{1}{20}=frac{19}{20}text{.})

(ds y=-frac{x}{2} 1text{.})

Find the beeline approximation of the action (f(x)=sqrt{1 x}) at (a=3text{,}) and use it to appraisal the amount of (sqrt{5}text{.}) Use a account to explain if this is an aggrandize or belittle of the absolute value.

(ds sqrt{5}approx L(4) = 2.25text{.}) Overestimate.

Let (f(x)=sqrt{1 2x}text{.})

Find the beeline approximation of (f(x)) at (x=0text{.})

Use your acknowledgment to appraisal the amount of (sqrt{1.1}text{.})

Is your appraisal an over- or under-estimate?

(L(x)=1 xtext{.})

(ds sqrt{1.1}=f(0.05)approx L(0.05)=1.05text{.})

An over-estimate back (f) is concave-down. MAPLE gives (sqrt{1.1} approx 1.048808848text{.})

Let (f(x)=sqrt[3]{x 8}text{.})

Find a beeline approximation to the action (f(x)=sqrt[3]{x 8}) at (a=0text{.})

Use this approximation to appraisal the numbers (sqrt[3]{7.95}) and (sqrt[3]{8.1}text{.})

(ds L(x)=2 frac{x}{12}text{.})

(ds sqrt[3]{7.95}approx L(-0.05)=2-frac{1}{240}=frac{479}{240}) and (ds sqrt[3]{8.1}approx L(0.1)=2 frac{1}{120}=frac{243}{120}text{.}) (Note: MAPLE gives (ds frac{479}{240}approx 1.995833333) and (sqrt[3]{7.95}approx 1.995824623text{.}) Also, (ds frac{243}{120}approx 2.025000000) and (sqrt[3]{8.1}approx 2.008298850text{.}))

Let (f(x)=(1 x)^{100}text{.})

Construct the beeline approximation to (f(x)=(1 x)^{100}text{.})

Use your approximation from (a) to appraisal ((1.0003)^{100}text{.})

Is your appraisal from (b) college or lower than the accurate value? Explain.

For (ain mathbb{R}text{,}) (L(x) = (1 a)^{100} 100(1 a)^{99}(x-a)text{.})

((1.0003)^{100} approx L(0.0003) = 1.03text{.})

Underestimate.

Let (f(x)=sqrt[3]{27 3x}text{.})

Find the blueprint of the departure band to the blueprint of the action (f(x)=sqrt[3]{27 3x}) at (x=0text{.})

Use your acknowledgment to appraisal a amount of (sqrt[3]{30}text{.})

Draw a account to appearance how the blueprint of (f) and its departure band behave about the point area (x=0) and the amount of (x) area the amount in allotment (b) is obtained.

(ds y=frac{x}{9} 3text{.})

(ds sqrt[3]{30}approx frac{1}{9} 3=frac{28}{9}text{.}) (Note: MAPLE gives (ds frac{28}{9}approx 3.111111111) and (sqrt[3]{30}approx 3.107232506text{.}))

Use beeline approximation to appraisal the amount of (ln 0.9text{.})

(ln 0.9approx -0.1text{.})

The linearization of the action (f(x)=ln x) at (a=1) is accustomed by (L(x)=x-1text{.}) Thus (ln 0.9approx L(0.9)=-0.1text{.}) (Note: MAPLE gives (ln 0.9approx -.1053605157text{.}))

Use a beeline approximation to appraisal the amount of (ds e^{-0.015}text{.}) Is your appraisal too ample or too small?

Take (f(x)=e^xtext{.})

(e^{-0.015} approx 0.985text{.}) Underestimate.

Let (f(x)=ln xtext{.})

Write the beeline approximation for (f(x)=ln x) about 1.

Compute the approximated amount for (exp (-0.1)) application beeline approximation.

(L(x)=x-1text{.})

Let (x=exp (-0.1)text{.}) Again (ln x=-0.1approx L(x)=x-1text{.}) Thus (xapprox 0.9text{.}) (Note: MAPLE gives (exp (-0.1)approx 0.9048374180text{.}))

Using the action (f(x)=x^{1/3}) and the address of beeline approximation, accord an appraisal for (1001^{1/3}text{.})

(ds L(x)=10 frac{1}{300}(x-1000)) implies (ds 1001^{1/3}approx L(1001)=frac{3001}{300}text{.}) (Note: MAPLE gives (ds frac{3001}{300}approx 10.00333333) and (sqrt[3]{1001}approx 10.00333222text{.}))

Let (ds f(x)=sqrt{x} sqrt[5]{x}text{.})

Use beeline approximation to actuate which of the afterward is abutting the amount of (f(1.001)text{:})

begin{equation*} begin{array}{lllll} 2.0001amp 2.0002amp 2.0003amp 2.0005amp 2.0007\ 2.001amp 2.002amp 2.003amp 2.005amp 2.007 end{array} end{equation*}

At (x=1text{,}) is (f(x)) biconcave up or biconcave down?

Based on your acknowledgment above, is your appraisal of (f(1.001)) too aerial or too low?

The linearization of the action (ds f(x)=sqrt{x} sqrt[5]{x}) at (a=1) is accustomed by (ds L(x)=2 frac{7}{10}(x-1)text{.}) Thus (f(1.001)approx L(1.001)=2 0.7cdot 0.001=2.0007text{.})

Note that the area of the action (f) is the breach ([0,infty )text{.}) From (ds f”(x) =-frac{1}{4}x^{-frac{3}{2}}-frac{4}{25}x^{-frac{9}{5}}) it follows that (f) is biconcave downwards on the breach ((0,infty)text{.})

The blueprint of the action is beneath the departure band at (a=1text{,}) so the appraisal (f(1.001)approx 2.0007) is too high.

Find the beeline approximation of (f(x)=sin x) about the point (x=pi /6text{.})

Explain why (f) satisfies the altitude of the Mean Amount Theorem. Use the assumption to prove that (ds sin xleq frac{1}{2} left(x-frac{pi }{6}right)) on the breach (ds left[frac{pi }{6},xright]) area (ds x>frac{pi }{6}text{.})

Is the cogwheel (df) beyond or abate than (Delta f) on the breach (ds left[frac{pi }{6},xright]text{?}) Do not accomplish any calculations. Use alone the after-effects in allotment (a) and (b) to explain your answer.

(ds L(x)=frac{1}{2} frac{sqrt{3}}{2}left(x-frac{pi }{6}right)text{.})

By the Mean Amount Theorem, for (ds x>frac{pi }{6}) and some (ds cin left(frac{pi }{6},xright)text{,}) (ds frac{f(x)-f(frac{pi }{6})}{x-frac{pi }{6}}=frac{sin x-frac{1}{2}}{x-frac{pi }{6}}=f^prime(c)=cos cleq 1text{.}) Back (ds x-frac{pi }{6}>0text{,}) the asperity follows.

From (a) and (b) it follows that, for (ds x>frac{pi }{6}text{,}) (ds sin xleq frac{1}{2} (x-frac{pi }{6})lt frac{1}{2} frac{sqrt{3}}{2}(x-frac{pi }{6})=L(x)text{.}) Next, (Delta f=f(x)-f(frac{pi }{6})=sin x-frac{1}{2}lt L(x)-frac{1}{2}=frac{sqrt{3}}{2}(x-frac{pi }{6})= f^prime (frac{pi }{6})Delta x=dftext{.})

Suppose that the alone advice we accept about a action (f) is that (f(1)=5) and that the blueprint of its acquired is as apparent in the Figure.

Use a beeline approximation to appraisal (f(0.9)) and (f(1.1)text{.})

Are your estimates in allotment (a) too ample or too small?

(f(0.9) approx L(0.9) = 5.2text{,}) (f(1.1) approx L(1.1) = 4.8text{.})

Too large.

Suppose that the alone advice we accept about a action (f) is that (f(1)=3) and that the blueprint of its acquired is as apparent in the Figure.

Use a beeline approximation to appraisal (f(0.9)) and (f(1.1)text{.})

Are your estimates in allotment (a) too ample or too small?

(f(0.9) approx L(0.9) = 2.8text{,}) (f(1.1) approx L(1.1) = 3.2text{.})

Too large.

State Newton’s accepted blueprint that produces a arrangement of approximations (x_1,x_2, x_3, ldots) to a basis of action (f(x)text{.})

Find the absolute basis of the blueprint (cos x=x^2) application Newton’s method, actual to 3 decimal points, with the aboriginal approximation (x_1=1text{.})

Let (f(x)=cos x-x^2text{.}) Again (f'(x)=-sin x-2xtext{.}) Thus (ds x_2=1-frac{cos 1-1}{-sin 1-2}approx 0.8382184099text{,}) (ds x_2=0.8382184099-frac{cos 0.8382184099-0.8382184099^2}{-sin 0.8382184099-2cdot 0.8382184099}approx \ 0.8242418682text{,}) and (ds x_3=0.8242418682-frac{cos 0.8242418682-0.8242418682^2}{-sin 0.8242418682-2cdot 0.8242418682}approx \ 0.8241323190text{.}) (Note: MAPLE gives (cos 0.8241323190-0.8241323190^2approx -1.59cdot 10^{-8}text{.}))

State Newton’s accepted blueprint that produces a arrangement of approximations (x_0,x_1,x_2, ldots) to a band-aid of (f(x)=0text{,}) bold that (x_0) is given.

Draw a labeled diagram assuming an archetype of a action (f(x)) for which Newton’s accepted blueprint fails to acquisition a band-aid of (f(x)=0text{.}) Mark on your diagram (x_0text{,}) (x_1text{,}) and (x_2text{.})

Take (f(x)=sqrt[3]{x}text{,}) (x_0=1text{,}) (x_1=-2text{,}) (x_2=4text{,}) and (x_3=-8text{.})

Explain how you can use Newton’s Adjustment to almost the amount of (sqrt{5}text{.})

Explain which of the afterward choices is the best antecedent approximation back application Newton’s Adjustment as in (a):(-1text{,}) (0text{,}) or (1text{?})

Find the fourth approximation (x_4) to the amount of (sqrt{5}) application Newton’s Adjustment with the antecedent approximation (x_1) you chose in (b).

We use Newton’s Adjustment to break the blueprint (x^2-5=0text{,}) (x>0text{.}) From (f(x)=x^2-5) and (f^prime (x)=2xtext{,}) Newton’s Adjustment gives (ds x_{n 1}=x_n-frac{x_n^2-5}{2x_n}=frac{1}{2}left( x_n frac{5}{x_n}right)text{.})

A asperous appraisal of (sqrt{5}) gives a amount that is a bit bigger than 2. Thus, booty (x_1=1text{.})

(ds x_2=3text{,}) (ds x_3=frac{7}{3}text{,}) (x_4=frac{47}{21}approx 2.23809text{.}) (Note: MAPLE gives (sqrt{5}approx 2.23606text{.}))

Apply Newton’s adjustment to (ds f(x)=x^{1/3}) with (x_0=1) and account (x_1,x_2,x_3, x_4text{.}) Acquisition a blueprint for (|x_n|text{.}) What happens to (|x_n|) as (nto inftytext{?}) Draw a account that shows what is activity on.

Let (ds f(x)=x^{frac{1}{3}}text{.}) Again Newton’s adjustment gives (displaystyle x_{n 1}=x_n-frac{f(x_n)}{f,'(x_n)}= x_n-frac{x_n^{frac{1}{3}}}{frac{1}{3}x_n^{-frac{2}{3}}} = -2x_ntext{.}) So (|x_{n 1}|=2|x_n|text{.}) This implies that if (x_0neq0, |x_n| = 2^n|x_0|rightarrowinfty) as (nrightarrowinftytext{;}) Newton’s Adjustment does not assignment in this case!

Find the Newton’s adjustment abundance blueprint to appraisal (sqrt[3]{68}text{.})

Provide an antecedent guess. Again explain, whether your antecedent assumption is an over- or belittle afterwards the aboriginal iteration.

(ds x_{n 1}=x_n-frac{x_n^3-68}{3x_n^2}text{.})

(x_0=4text{.}) Underestimate.

Use beeline approximation to appraisal (sqrt[3]{26}text{.})

The amount of (sqrt[3]{26}) is about (x_1=3text{.}) Use Newton’s adjustment to acquisition a bigger approximation, (x_2text{,}) to (sqrt[3]{26}text{.})

(sqrt[3]{26}approx L(26)approx 2.962text{.})

Take (f(x)=x^3-26text{.}) (x_1=3) implies (x_2approx 2.962text{.})

This catechism apropos award zeros of the function

begin{equation*} f(x)=left{ begin{array}{rr} sqrt{x}amp mbox{if } xgeq 0\ -sqrt{-x}amp mbox{if } xlt 0. end{array} right. end{equation*}

If the antecedent approximation to the zero, for (f(x)) accustomed above, is (x_1text{,}) what blueprint does Newton’s adjustment accord for the abutting approximation?

The basis of the blueprint (f(x)=0) is (x=0text{.}) Explain why Newton’s adjustment fails to acquisition the basis no amount which antecedent approximation (x_1not=0) is used. Allegorize your account with a sketch.

Observe that (ds f(x)=mbox{sign} (x)cdot sqrt{|x|}) and (ds f^prime(x)=frac{1}{2sqrt{|x|}}text{,}) so (x_{n 1}=-x_ntext{.})

By induction, for all (nin mathbb{N}text{,}) (|x_{n}|=|x_1|text{.})

Suppose (k) is a constant. Appearance that if we administer Newton’s adjustment to almost the amount of (sqrt[5]{k}text{,}) we get the afterward accepted formula:

begin{equation*} x_{n 1}=frac{x_n}{5}left( 4 frac{k}{x_n^5}right)text{.} end{equation*}

If (x_n=sqrt[5]{k}text{,}) what is the amount of (x_{n 1}text{?})

Take (x_1=2) and use the blueprint in allotment (a) to acquisition (x_2text{,}) an appraisal of the amount of (sqrt[5]{20}) that is actual to one decimal place.

Take (f(x)=x^5-ktext{.}) Again (f^prime (x)=5x^4) and (ds x_{n 1}=x_n-frac{x_n^5-k}{5x_n^4}=frac{4x_n^5 k}{5x_n^4}=frac{x_n}{5}left( 4 frac{k}{x_n^5}right)text{.})

(ds x_{n 1}=sqrt[5]{k}text{.})

(x_2=1.85text{.}) [MAPLE gives (sqrt[5]{20}approx 1.820564203text{.})]

Use Newton’s adjustment to acquisition the additional approximation (x_2) of (sqrt[5]{31}) starting with the antecedent approximation (x_0=2text{.})

From (f(x)=x^5-31) and (f'(x)=5x^4) it follows that (ds x_1=frac{159}{80}) and (ds x_2=frac{159}{80}-frac{left( frac{159}{80}right) ^5-31}{5cdot left( frac{159}{80}right) ^4}approx 1.987340780text{.}) (Note: MAPLE gives (sqrt[5]{31}=1.987340755text{.}))

Suppose (x_0) is an antecedent appraisal in Newton’s adjustment activated to the action (f(x)text{.}) Acquire Newton’s blueprint for (x_1text{,}) namely

begin{equation*} x_1=x_0-frac{f(x_0)}{f'(x_0)}text{.} end{equation*}

Support your ancestry with a account assuming a action (f(x)text{,}) with (x_0text{,}) (x_1) and the band whose abruptness is (f'(x_0)) acutely labeled.

Using one abundance of Newton’s adjustment with (ds x_0=frac{pi }{2}) almost the (x)-coordinate of the point area the action (g(x)=sin x) crosses the band (y=xtext{.})

The catechism is almost the band-aid of the blueprint (F(x)=sin x-x=0) with (ds x_0=frac{pi }{2}text{.}) Thus (ds x_1=frac{pi }{2}-frac{sin frac{pi }{2}-frac{pi }{2}}{cos frac{pi }{2}-1}=1text{.}) (Note: Acutely the band-aid of the accustomed blueprint is (x=0text{.}) Newton’s adjustment with (x_0=frac{pi }{2}) gives (x_7=0.08518323251text{.}))

The equation

begin{equation*} 8x^3-12x^2-22x 25=0 end{equation*}

has a band-aid abreast (x_1=1text{.}) Use Newton’s Adjustment to acquisition a bigger approximation (x_2) to this solution. Express your acknowledgment as a fraction.

(ds x_2=1-frac{-1}{-22}=frac{21}{22}text{.}) Note: MAPLE gives (ds frac{21}{22}approx 0.9545454545) and approximates the band-aid of the blueprint as (0.9555894038))

The departure band to the blueprint (y=f(x)) at the point (A(2,-1)) is accustomed by (y=-1 4(x-2)text{.}) It is additionally accepted that (f^{primeprime}(2)=3text{.})

Assume that Newton’s Adjustment is acclimated to break the blueprint (f(x)=0) and (x_0=2) is the antecedent guess. Acquisition the abutting approximation, (x_1text{,}) to the solution.

Assume that Newton’s Adjustment is acclimated to acquisition a analytical point for (f) and that (x_0=2) is the antecedent guess. Acquisition the abutting approximation, (x_1text{,}) to the analytical point.

(ds x_1=2-frac{-1}{4}=frac{9}{4}text{.})

The catechism is to almost a band-aid of the blueprint (f'(x)=0) with the antecedent assumption (x_0=2text{,}) (f'(2)=4text{,}) and (f”(2)=3) given. Hence (x_1=2-frac{4}{3}=frac{2}{3}text{.})

Apply Newton’s adjustment to the blueprint (ds frac{1}{x}-a=0) to acquire the afterward algorithm for award reciprocals:

begin{equation*} x_{n 1}=2x_n-ax_n^2text{.} end{equation*}

Use the algorithm from allotment (a) to account (ds frac{1}{1.128}) actual to three decimal places, starting with the aboriginal approximation (x_1=1text{.})

From (ds f(x)=frac{1}{x}-a) and (ds f'(x)=-frac{1}{x^2}) it follows that (ds x_{n 1}=x_n-frac{frac{1}{x_n}-a}{-frac{1}{x_n^2}}=2x_n-ax_n^2text{.})

Note that (ds frac{1}{1.128}) is the band-aid of the blueprint (ds frac{1}{x}-1.128=0text{.}) Thus (x_2=2-1.128=0.872text{,}) (x_3=2cdot 0.872-1.128cdot 0.872^2=0.886286848text{,}) and (x_4=0.8865247589text{.}) (Note: MAPLE gives (ds frac{1}{1.128}approx 0.8865248227text{.}))

Apply Newton’s adjustment to the blueprint (ds x^2-a=0) to acquire the afterward algorithm for the roots:

begin{equation*} x_{n 1}=frac{1}{2}left(x_n frac{a}{x_n}right)text{.} end{equation*}

Approximate (sqrt{2}) by demography (x_1=2) and artful (x_2text{.})

(ds x_2=frac{3}{2}text{.})

State the blueprint for the linearization of (f) at (atext{,})

Using beeline approximation, almost (sqrt[4]{81.1}text{.})

Approximate (sqrt[4]{81.1}) application one abundance of Newton’s method.

Take (f(x)=sqrt[4]{x}) with (a=81text{.}) (sqrt[4]{81.1} approx L(81.1)approx 3.000925text{.})

Take (g(x)=x^4-81.1text{.}) (x_2= 3.000925text{.})

You seek the almost amount of (x) which is abreast 1.8 for which (ds sin x=frac{x}{2}text{.}) Your aboriginal assumption is that (xapprox x_1=frac{pi }{2}text{.}) Use one abundance of Newton’s adjustment to acquisition a bigger approximation to (xtext{.}) Abridge your acknowledgment as far as possible.

(ds x_2=frac{pi }{2}-frac{1-frac{pi }{4}}{-frac{1}{2}}=2text{.}) (Note: MAPLE estimates the absolute band-aid of the blueprint (ds sin x=frac{x}{2}) as (1.895494267text{.}) Newton’s adjustment with the antecedent assumption (ds x_1=frac{pi }{2}) gives (x_3approx 1.900995594text{.}))

For the action (f(x)=x^3-3x 5) use the Intermediate Amount Theorem, and any added accoutrement you charge to actuate intervals of breadth 1 anniversary of which contains a basis of (ftext{.})

Pick one of the intervals begin in allotment (a). Choose the larboard endpoint of this breach to be (x_0text{.}) Now, use this as a starting amount to acquisition two new iterations to the basis of (f) by application Newton’s method. Actuate from these whether Newton’s adjustment is working. Absolve your acknowledgment carefully.

From (f'(x)=3(x^2-1)) it follows that the analytical numbers are (x=pm 1text{.}) From (f(1)=3text{,}) (f(-1)=7text{,}) (ds lim _{xto -infty}f(x)=-inftytext{,}) and (ds lim _{xto infty}f(x)=infty) it follows that (f) has alone one basis and that basis belongs to the breach ((-infty ,-1)text{.}) From (f(-2)=3>0) and (f(-3)=-13lt 0text{,}) by the Intermediate Amount Theorem, we achieve that the basis belongs to the breach ((-3,-2)text{.})

Let (x_0=-3text{.}) Again (ds x_1=-3-frac{-13}{504}=-frac{1499}{504}approx -2.974206349) and (x_3approx -2.447947724text{.}) It seems that Newton’s adjustment is working, the new iterations are central the breach ((-3,-2)) area we apperceive that the basis is. (Note: MAPLE estimates the band-aid of the blueprint (ds x^3-3x 5=0) as (x=-2.279018786text{.}))

Let (f(x)=x^3 3x 1text{.})

Show that (f) has at atomic one basis in the breach (ds left( -frac{1}{2},0right)text{.}) Explain your reasoning.

Use Newton’s adjustment to almost the basis that lies in the breach (ds left( -frac{1}{2},0right)text{.}) Stop back the abutting abundance agrees with the antecedent one to two decimal places.

The action (f) is connected on the bankrupt breach (ds left[ -frac{1}{2},0right]) and (ds fleft( -frac{1}{2}right)=-frac{5}{8}lt 0) and (f(0)=1>0text{.}) By the Intermediate Amount Theorem, the action (f) has at atomic one basis in the breach (ds left( -frac{1}{2},0right)text{.})

Take (ds x_1=-frac{1}{3}text{.}) Again (ds x_2=-frac{1}{3}-frac{-frac{1}{27}-3cdot left( -frac{1}{3}right) 1}{3left( frac{1}{9} 1right)} =-frac{29}{90}approx -.3222222222) and (ds x_3approx -.3221853550text{.}) (Note: MAPLE estimates the band-aid of the blueprint (ds x^3 3x 1=0) as (x=-.3221853546text{.}))

In this catechism we investigate the band-aid of the blueprint (ln x=-x^2 3) on the breach ([1,3]text{.})

Explain why you apperceive the blueprint has at atomic one band-aid on ([1,3]text{.})

Show that the blueprint has absolutely one band-aid on ([1,3]text{.})

Use Newton’s Adjustment to almost the band-aid of the blueprint by starting with (x_1=1) and award (x_2text{.})

Take (f(x)=ln x x^2-3text{,}) appraise (f(1)) and (f(3)text{,}) and again use the Intermediate Amount Theorem.

Note that (ds f^prime (x)=frac{1}{x} 2x>0) for (xin (1,3)text{.})

From (ds f(1)=-2) and (f^prime (1)=3) it follows that (ds x_2=frac{5}{3}approx 1.66text{.}) [MAPLE gives 1.592142937 as the solution.]

In this catechism we investigate the absolute band-aid of the blueprint (x^2 x=5-ln xtext{.})

Explain why you apperceive the blueprint has at atomic one absolute solution.

Show that the blueprint has absolutely one absolute solution.

Use Newton’s Adjustment to almost the band-aid of the blueprint by starting with (x_1=1) and award (x_2text{.})

Use the Intermediate Amount Theorem.

Use Rolle’s Theorem.

(x_2=1.75text{.})

In this catechism we investigate the band-aid of the blueprint (2x=cos xtext{.})

Explain why you apperceive the blueprint has at atomic one solution.

Show that the blueprint has absolutely one solution.

Use Newton’s Adjustment to almost the band-aid of the blueprint by starting with (x_1=0) and award (x_2text{.})

Take (f(x)=2x-cos xtext{,}) appraise (ds lim _{xto -infty }f(x)) and (ds lim _{xto infty }f(x)text{,}) and again use the Intermediate Amount Theorem.

Note that (ds f^prime (x)=2 sin x>0) for (xin mathbb{R}text{.})

From (ds f(0)=-1) and (f^prime (0)=2) it follows that (ds x_2=frac{1}{2}text{.}) [MAPLE gives 0.4501836113 as the solution.]

In this catechism we investigate the band-aid of the blueprint (2x-1=sin xtext{.})

Explain why you apperceive the blueprint has at atomic one solution.

Show that the blueprint has absolutely one solution.

Use Newton’s Adjustment to almost the band-aid of the blueprint by starting with (x_1=0) and award (x_2text{.})

Take (f(x)=2x-1-sin xtext{,}) appraise (ds lim _{xto -infty }f(x)) and (ds lim _{xto infty }f(x)text{,}) and again use the Intermediate Amount Theorem.

Note that (ds f^prime (x)=2-cos x>0) for (xin mathbb{R}text{.})

From (ds f(0)=-1) and (f^prime (0)=1) it follows that (ds x_2=1text{.}) [MAPLE gives 0.8878622116 as the solution.]

In this catechism we investigate the absolute band-aid of the blueprint (e^x=2cos xtext{.})

Explain why you apperceive the blueprint has at atomic one absolute solution.

Show that the blueprint has absolutely one absolute solution.

Use Newton’s Adjustment to almost the band-aid of the blueprint by starting with (x_1=0) and award (x_2text{.})

Take (f(x)=e^x-cos 2x) and again use the Intermediate Amount Theorem.

Observe that (f^prime(x)>0) for all (x>0text{.})

(x_2=1text{.})

Consider the equation

begin{equation*} x^6-x-1=0text{.} end{equation*}

Apply the Intermediate Amount Assumption to the action (f(x)=x^6-x-1) to prove that the accustomed blueprint has a basis greater than 1. Make abiding that you absolve why the action (f) is connected on its domain.

Use the acquired of the action (f(x)=x^6-x-1) to prove that the accustomed blueprint has alone one basis greater than 1. Call that basis (atext{.}) Appearance all your work. Acutely explain your reasoning.

State Newton’s Method.

Use Newton’s Adjustment with the antecedent approximation (x_1=1) to acquisition (x_2) and (x_3text{,}) the additional and the third approximations to the basis (a) of the blueprint (x^6-x-1=0text{.}) You may use your calculator to acquisition those values. Appearance all your work. Acutely explain your reasoning.

WolframAlpha gives (aapprox 1.13472text{.}) Use your calculator to appraise the cardinal (|x_3 -1.13472|text{.}) Are you annoyed with your approximation. Why yes or why not?

Take (f(x)=x^6-x-1) and again use the Intermediate Amount Theorem.

(x_2=1.2) and (x_3approx 1.143575843text{.})

(|x_3-1.1347|approx 0.008855843text{.})

State Rolle’s theorem.

Use Rolle’s assumption to prove that (f(x)) has a analytical point in ([0,1]) where

begin{equation*} f(x)=sinleft(frac{pi x}{2}right) -x^2text{.} end{equation*}

Set up the Newton’s adjustment abundance blueprint ((x_{n 1}) in agreement of (x_n)) to almost the analytical point You do not charge to simplify.

(ds x_{n 1}=x_n-frac{sinleft(frac{pi x_n}{2}right)-x_n^2}{frac{pi}{2}cosleft(frac{pi x_n}{2}right)-2x_n}text{.})

State the Mean Amount Theorem.

Using the Mean Amount Theorem, prove that (f(x)) has a analytical point in ([0,1]) where

begin{equation*} f(x)=cosleft(frac{pi x}{2}right) xtext{.} end{equation*}

Set up the Newton’s adjustment abundance blueprint to almost the analytical point You do not charge to simplify.

(ds x_{n 1}=x_n-frac{cosleft(frac{pi x_n}{2}right) x_n}{frac{-pi}{2}cosleft(frac{pi x_n}{2}right) 1}text{.})

State the Intermediate Amount Theorem.

State the Mean Amount Theorem.

Use the Intermediate Amount Assumption and the Mean Amount Assumption to appearance that the blueprint (1 2x x^3 4x^5=0) has absolutely one absolute root.

Proof by contradiction.

A action (h(x)) is said to accept a anchored point at (x=c) if (h(c)=ctext{.}) Suppose that the area and ambit of a action (f(x)) are both the breach ([0,1]) and that (f) is connected on this domain, with (f(0)not= 0) and (f(1)not=1text{.})

Prove that (f) has at atomic one anchored point. That is, prove that (f(c)=c) for some (cin(0,1)text{.})

Suppose that (f^prime(x)lt 1) for all (xin(0,1)text{.}) Prove that (f) has absolutely one anchored point in ([0,1]text{.})

Use Newton’s adjustment to actuate an abundance blueprint for the anchored point (x=ctext{.})

Consider (g(x)=f(x)-xtext{.})

(ds x_{n 1}=x_n-frac{f(x_n)-x_n}{f^prime(x_n)-1}text{.})

How To Write An Absolute Value Equation From A Graph – How To Write An Absolute Value Equation From A Graph

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