Solved on Sep 28, 2023

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

STEP 1

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

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

Rearrange the equation and solve for $x$ .
$27x^2 =108$ $x^2 =4$ $x = \pm2$

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

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

STEP 15

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

STEP 16

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

STEP 17

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

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

STEP 1

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

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

Rearrange the equation and solve for $x$ .
$27x^2 =108$ $x^2 =4$ $x = \pm2$

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

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

STEP 15

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

STEP 16

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

STEP 17

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

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STEP 1

STEP 2

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

STEP 3

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

STEP 4

Now, we need to find the derivative of the function f(x)f(x)f(x).f′(x)=d(9x3−90x+157)/dxf'(x) = d(9x^3 -90x +157)/dxf′(x)=d(9x3−90x+157)/dx

STEP 5

Calculate the derivative of f(x)f(x)f(x).f′(x)=27x2−90f'(x) =27x^2 -90f′(x)=27x2−90

STEP 6

Set the derivative of f(x)f(x)f(x) equal to the derivative of (x)(x)(x) and solve for xxx.27x2−90=1827x^2 -90 =1827x2−90=18

STEP 7

Rearrange the equation and solve for xxx.27x2=10827x^2 =10827x2=108x2=4x^2 =4x2=4x=±2x = \pm2x=±2

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

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

STEP 15

Check the yyy-intercept of the function g(x)g(x)g(x) at x=0.2x=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) +22g(0.2)=9ln(5(0.2))−27(0.2)+22

STEP 16

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

STEP 17

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

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

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

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

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

Rearrange the equation and solve for $x$ .
$27x^2 =108$ $x^2 =4$ $x = \pm2$

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

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

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

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

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

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

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

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

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