Question

let $(x_0,y_0)=(3,6)$ and $(x_1,y_1)=(3.5,6.2)$. use the following graph of the function $f$ to find the indicated derivatives.
if $h(x)=(f(x))^4$, then
$h(3)=$
if $g(x)=f^{-1}(x)$, then
$g(6)=$

Explanation:

Step1: Apply chain - rule for $h(x)=(f(x))^{4}$

The chain - rule states that if $h(x)=(f(x))^{4}$, then $h'(x) = 4(f(x))^{3}\cdot f'(x)$. To find $h'(3)$, we need to know $f(3)$ and $f'(3)$. From the given points, when $x = 3$, $y=f(3)=6$. The slope of the secant line between $(x_0,y_0)=(3,6)$ and $(x_1,y_1)=(3.5,6.2)$ can be used as an approximation for $f'(3)$. The slope of the secant line $m=\frac{y_1 - y_0}{x_1 - x_0}=\frac{6.2 - 6}{3.5 - 3}=\frac{0.2}{0.5}=0.4$. So $h'(3)=4(f(3))^{3}\cdot f'(3)$. Substituting $f(3) = 6$ and $f'(3)\approx0.4$, we get $h'(3)=4\times6^{3}\times0.4$.

Step2: Calculate the value of $h'(3)$

$h'(3)=4\times216\times0.4=4\times86.4 = 345.6$.

Step3: Use the formula for the derivative of an inverse function

If $g(x)=f^{-1}(x)$, then $g'(x)=\frac{1}{f'(g(x))}$. We want to find $g'(6)$. Since $f(3)=6$, then $g(6)=3$. So $g'(6)=\frac{1}{f'(3)}$. We found that $f'(3)\approx0.4$, so $g'(6)=\frac{1}{0.4}=2.5$.

Answer:

$h'(3)=345.6$
$g'(6)=2.5$