Question

given the function below

f(x)=sqrt3{-50 x^{3}+175}

find the equation of the tangent line to the graph of the function at ( x = 1 ). answer in ( m x + b ) form.

l(x)=

use the tangent line to approximate ( f(1.1) ).

l(1.1)=

compute the actual value of ( f(1.1) ). what is the error between the function value and the linear approximation?
answer as a positive value only.

| \text { error } | approx (approximate to at least 5 decimal places.)

question help: video message instructor

Explanation:

Step1: Find \( f(1) \)

Substitute \( x = 1 \) into \( f(x)=\sqrt[3]{-50x^{3}+175} \).
\( f(1)=\sqrt[3]{-50\times1^{3}+175}=\sqrt[3]{125} = 5 \)

Step2: Find the derivative \( f^{\prime}(x) \)

Let \( u=-50x^{3}+175 \), then \( y = u^{\frac{1}{3}} \).
Using the chain - rule \( y^{\prime}=\frac{dy}{du}\cdot\frac{du}{dx} \).
\( \frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}} \) and \( \frac{du}{dx}=-150x^{2} \).
So \( f^{\prime}(x)=\frac{-150x^{2}}{3(-50x^{3}+175)^{\frac{2}{3}}}=\frac{-50x^{2}}{(-50x^{3}+175)^{\frac{2}{3}}} \)

Step3: Find \( f^{\prime}(1) \)

Substitute \( x = 1 \) into \( f^{\prime}(x) \).
\( f^{\prime}(1)=\frac{-50\times1^{2}}{(-50\times1^{3}+175)^{\frac{2}{3}}}=\frac{-50}{125^{\frac{2}{3}}}=\frac{-50}{25}=-2 \)

Step4: Find the equation of the tangent line \( L(x) \)

Using the point - slope form \( y - y_{0}=m(x - x_{0}) \), where \( (x_{0},y_{0})=(1,5) \) and \( m=-2 \).
\( y - 5=-2(x - 1) \)
\( y=-2x + 2+5 \)
\( L(x)=-2x + 7 \)

Step5: Approximate \( f(1.1) \) using \( L(x) \)

Substitute \( x = 1.1 \) into \( L(x) \).
\( L(1.1)=-2\times1.1 + 7=-2.2 + 7 = 4.8 \)

Step6: Compute the actual value of \( f(1.1) \)

\( f(1.1)=\sqrt[3]{-50\times(1.1)^{3}+175}=\sqrt[3]{-50\times1.331 + 175}=\sqrt[3]{-66.55+175}=\sqrt[3]{108.45}\approx4.76797 \)

Step7: Compute the error

\( |\text{error}|=|4.8 - 4.76797|=0.03203 \)

Answer:

\( L(x)=-2x + 7 \)
\( L(1.1)=4.8 \)
\( |\text{error}|\approx0.03203 \)