Let f(x)=x^5+2e^x/4 for all x R.

Question

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

Accepted Answer

Given that  $$(g \circ f)(x) = x$$  for all $$x \in \mathbf{R}$$. This means $g(f(x)) = x$ for all $$x \in \mathbf{R}$$. Differentiating both sides with respect to $x$, we get:

$$g'(f(x)) \cdot f'(x) = 1$$

Now, we want to find the value of $8g'(2)$. To do this, we need to find a value of $x$ such that $f(x) = 2$. Let's solve for $x$:

$x^5 + 2e^{x/4} = 2$

By inspection, we see that $x = 0$ is a solution. Therefore, $f(0) = 2$. Now, we can substitute this into our differentiated equation:

$$g'(f(0)) \cdot f'(0) = 1$$

$$g'(2) \cdot f'(0) = 1$$

Let's find $f'(0)$:

$$f'(x) = 5x^4 + \frac{1}{2}e^{x/4}$$

$$f'(0) = \frac{1}{2}$$

Substituting this back into our equation:

$$g'(2) \cdot \frac{1}{2} = 1$$

$g'(2) = 2$

Finally, we can calculate $8g'(2)$:

$$8g'(2) = 8 \cdot 2 = \boxed{16}$$

Solutions · 1