Question

let (f) be a differentiable function with (f(1)= - 2) and (f^{prime}(1)=2). let the function (g(x)=2|f(x)|^{3}). write the equation of the line tangent to the graph of (g) at the point where (x = 1).

Explanation:

Step1: Find $g(1)$

Substitute $x = 1$ into $g(x)$. Since $f(1)=-2$, then $g(1)=2|f(1)|^{3}=2|-2|^{3}=2\times8 = 16$.

Step2: Differentiate $g(x)$ using chain - rule

Let $u = |f(x)|$. Then $g(x)=2u^{3}$. The derivative of $y = 2u^{3}$ with respect to $u$ is $\frac{dy}{du}=6u^{2}$. The derivative of $u = |f(x)|$ with respect to $x$ is $u^\prime=\frac{f(x)f^\prime(x)}{|f(x)|}$ (by the chain - rule for $|y|$ where $y = f(x)$). So $g^\prime(x)=6|f(x)|^{2}\cdot\frac{f(x)f^\prime(x)}{|f(x)|}=6f(x)f^\prime(x)$ for $f(x)
eq0$.

Step3: Find $g^\prime(1)$

Substitute $x = 1$ into $g^\prime(x)$. Given $f(1)=-2$ and $f^\prime(1)=2$, then $g^\prime(1)=6\times(-2)\times2=-24$.

Step4: Write the equation of the tangent line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,16)$ and $m = g^\prime(1)=-24$. So $y-16=-24(x - 1)$. Rearranging gives $y=-24x+40$.

Answer:

$y=-24x + 40$