Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by $f(x) = x^5 + x^3$ and let $g(x) = f^{-1}(x)$ be the inverse of $f$. If $g''(-2) = \frac{a}{b}$ where $a$ and $b$ are positive coprime integers, then the value of $a$ is

• A. 13
• B. 26
• C. 39
• D. 256

Hint:

Using the identity $f(g(x)) = x$ and differentiating implicitly twice is often easier to remember than the direct formula for the derivative of an inverse function.

Answer 1

The Correct Option is A

Step 1: Find the point on $f$ that maps to $-2$.
Since $g = f^{-1}$, the value $g(-2)$ is the number $x$ with $f(x) = -2$. Solve $x^5 + x^3 = -2$. Trying $x = -1$: $(-1)^5 + (-1)^3 = -1 - 1 = -2$. So $g(-2) = -1$.

Step 2: Recall the first derivative rule for inverses.
$g'(x) = \frac{1}{f'(g(x))}$. Here $f'(x) = 5x^4 + 3x^2$.

Step 3: Evaluate $f'$ at $g(-2) = -1$.
$f'(-1) = 5(1) + 3(1) = 8$. So $g'(-2) = \frac{1}{8}$.

Step 4: Recall the second derivative rule.
Differentiating gives $g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{[f'(g(x))]^2}$. We need $f''(x) = 20x^3 + 6x$.

Step 5: Evaluate $f''$ at $-1$.
$f''(-1) = 20(-1) + 6(-1) = -20 - 6 = -26$.

Step 6: Put the pieces together.
$g''(-2) = -\frac{(-26)\cdot \frac{1}{8}}{8^2} = -\frac{-26/8}{64} = \frac{26}{512} = \frac{13}{256}$. So $a = 13$ and $b = 256$ are coprime, giving $a = 13$.
\[ \boxed{a = 13} \]