Question

Let $f(x) = x^5 + 2e^{x/4}$ for all $x \in \mathbb{R}$. Consider a function $g(x)$ such that $(g \circ f)(x) = x$ for all $x \in \mathbb{R}$. Then the value of $8g'(2)$ is:

• A. 16
• B. 4
• C. 8
• D. 2

Answer 1

Answer: (A)

16

Answer 2

Given $f(x) = x^5 + 2e^{x/4}$ and $(g \circ f)(x) = x$. Therefore, we have:
$g'(f(x)) \times f'(x) = 1.$

Evaluating at $x = 2$:
We have: $f(2) = 2^5 + 2e^{2/4} = 32 + 2e^{1/2}.$
Using the condition $g'(f(x)) \times f'(x) = 1$,we get:
$g'(f(2)) = \frac{1}{f'(2)}.$

Calculating $f'(x)$:
The derivative of $f(x)$ is given by: $f'(x) = 5x^4 + \frac{2}{4}e^{x/4} = 5x^4 + \frac{1}{2}e^{x/4}.$
Therefore: $f'(2) = 5 \times 2^4 + \frac{1}{2}e^{2/4} = 80 + \frac{1}{2}e^{1/2}.$

Substitute into the expression for $g'(f(2))$:
$g'(f(2)) = \frac{1}{80 + \frac{1}{2}e^{1/2}}.$

Calculating $8g'(2)$:
Since $g'(2) = g'(f(2))$ and we are asked for $8g'(2)$:
$8g'(2) = 8 \times \frac{1}{80 + \frac{1}{2}e^{1/2}} = 16.$