Solved on Sep 28, 2023

Find the points x^0\hat{x}_{0} and x~0\tilde{x}_{0} where the linear function L(x)=18x+13L(x)=18x + 13 approximates f(x)=9x390x+157f(x)=9x^3 - 90x + 157 and g(x)=9ln(5x)27x+22g(x)=9\ln(5x) - 27x + 22, respectively. Compute the derivatives, set them equal to L(x)L'(x), and check the yy-intercept.

STEP 1

Assumptions1. The linear function is (x)=18x+13(x)=18x +13 . The functions f(x)=9x390x+157f(x)=9x^3 -90x +157 and g(x)=9ln(5x)27x+22g(x)=9\ln(5x) -27x +22 are differentiable3. The points x^0\hat{x}_{0} and x~0\tilde{x}_{0} are the points where the linear function is the linear approximation to the functions f(x)f(x) and g(x)g(x) respectively

STEP 2

First, we need to find the derivative of the linear function (x)(x).
(x)=d(18x+13)/dx'(x) = d(18x +13)/dx

STEP 3

Calculate the derivative of (x)(x).
(x)=18'(x) =18

STEP 4

Now, we need to find the derivative of the function f(x)f(x).
f(x)=d(9x390x+157)/dxf'(x) = d(9x^3 -90x +157)/dx

STEP 5

Calculate the derivative of f(x)f(x).
f(x)=27x290f'(x) =27x^2 -90

STEP 6

Set the derivative of f(x)f(x) equal to the derivative of (x)(x) and solve for xx.
27x290=1827x^2 -90 =18

STEP 7

Rearrange the equation and solve for xx.
27x2=10827x^2 =108x2=4x^2 =4x=±2x = \pm2

STEP 8

Check the yy-intercept of the function f(x)f(x) at x=2x=2 and x=2x=-2.
f(2)=(2)390(2)+157f(2) =(2)^3 -90(2) +157f(2)=(2)390(2)+157f(-2) =(-2)^3 -90(-2) +157

STEP 9

Calculate the values of f(2)f(2) and f(2)f(-2).
f(2)=9(8)180+157=29f(2) =9(8) -180 +157 =29f(2)=9(8)+180+157=301f(-2) =9(-8) +180 +157 =301

STEP 10

Since f(2)=L(2)f(2) = L(2) and f(2)L(2)f(-2) \neq L(-2), we conclude that x^0=2\hat{x}_{0} =2.

STEP 11

Now, we need to find the derivative of the function g(x)g(x).
g(x)=d(9ln(5x)27x+22)/dxg'(x) = d(9\ln(5x) -27x +22)/dx

STEP 12

Calculate the derivative of g(x)g(x).
g(x)=9/(x)27g'(x) =9/(x) -27

STEP 13

Set the derivative of g(x)g(x) equal to the derivative of (x)(x) and solve for xx.
9/x27=189/x -27 =18

STEP 14

Rearrange the equation and solve for xx.
9/x=459/x =45x=9/45=0.2x =9/45 =0.2

STEP 15

Check the yy-intercept of the function g(x)g(x) at x=0.2x=0.2.
g(0.2)=9ln(5(0.2))27(0.2)+22g(0.2) =9\ln(5(0.2)) -27(0.2) +22

STEP 16

Calculate the value of g(0.2)g(0.2).
g(0.2)=9ln()5.4+22=16.6g(0.2) =9\ln() -5.4 +22 =16.6

STEP 17

Since g(0.2)=L(0.2)g(0.2) = L(0.2), we conclude that x~0=0.2\tilde{x}_{0} =0.2.
The points where the linear function (x)=x+13(x)=x +13 is the linear approximation to the functions f(x)=9x390x+157f(x)=9x^3 -90x +157 and g(x)=9ln(5x)27x+22g(x)=9\ln(5x) -27x +22 are x^0=2\hat{x}_{0} =2 and x~0=0.2\tilde{x}_{0} =0.2, respectively.

Was this helpful?
banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ContactInfluencer programPolicyTerms
TwitterInstagramFacebookTikTokDiscord