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
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
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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.
Updated On: Jun 16, 2026
- 13
- 26
- 39
- 256
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Question:
The problem requires finding the second derivative of the inverse function $g(x) = f^{-1}(x)$ at a specific point $x = -2$.
Step 2: Key Formula or Approach:
We will use the derivative of inverse functions:
\[ g'(x) = \frac{1}{f'(g(x))} \]
Differentiating this relation again gives:
\[ g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{[f'(g(x))]^2} = -\frac{f''(g(x))}{[f'(g(x))]^3} \]
Step 3: Detailed Explanation:
• First, we need to find $g(-2)$, which is the value of $y$ such that $f(y) = -2$:
\[ f(y) = y^5 + y^3 = -2 \]
By inspection, we see that $y = -1$ is a solution:
\[ (-1)^5 + (-1)^3 = -1 - 1 = -2 \]
Since $f(x)$ is strictly increasing ($f'(x) = 5x^4 + 3x^2 \gt 0$ for $x \neq 0$), $y = -1$ is the unique real solution. Thus, $g(-2) = -1$.
• Now, we find the first and second derivatives of $f(x)$:
\[ f'(x) = 5x^4 + 3x^2 \]
\[ f''(x) = 20x^3 + 6x \]
• Next, evaluate these derivatives at $y = -1$:
\[ f'(-1) = 5(-1)^4 + 3(-1)^2 = 5 + 3 = 8 \]
\[ f''(-1) = 20(-1)^3 + 6(-1) = -20 - 6 = -26 \]
• Substitute these values into the formula for $g''(-2)$:
\[ g''(-2) = -\frac{f''(-1)}{[f'(-1)]^3} \]
\[ g''(-2) = -\frac{-26}{8^3} = \frac{26}{512} = \frac{13}{256} \]
• We are given $g''(-2) = \frac{a}{b}$ where $a$ and $b$ are positive coprime integers:
\[ a = 13, \quad b = 256 \]
Step 4: Final Answer:
The value of $a$ is 13.
The problem requires finding the second derivative of the inverse function $g(x) = f^{-1}(x)$ at a specific point $x = -2$.
Step 2: Key Formula or Approach:
We will use the derivative of inverse functions:
\[ g'(x) = \frac{1}{f'(g(x))} \]
Differentiating this relation again gives:
\[ g''(x) = -\frac{f''(g(x)) \cdot g'(x)}{[f'(g(x))]^2} = -\frac{f''(g(x))}{[f'(g(x))]^3} \]
Step 3: Detailed Explanation:
• First, we need to find $g(-2)$, which is the value of $y$ such that $f(y) = -2$:
\[ f(y) = y^5 + y^3 = -2 \]
By inspection, we see that $y = -1$ is a solution:
\[ (-1)^5 + (-1)^3 = -1 - 1 = -2 \]
Since $f(x)$ is strictly increasing ($f'(x) = 5x^4 + 3x^2 \gt 0$ for $x \neq 0$), $y = -1$ is the unique real solution. Thus, $g(-2) = -1$.
• Now, we find the first and second derivatives of $f(x)$:
\[ f'(x) = 5x^4 + 3x^2 \]
\[ f''(x) = 20x^3 + 6x \]
• Next, evaluate these derivatives at $y = -1$:
\[ f'(-1) = 5(-1)^4 + 3(-1)^2 = 5 + 3 = 8 \]
\[ f''(-1) = 20(-1)^3 + 6(-1) = -20 - 6 = -26 \]
• Substitute these values into the formula for $g''(-2)$:
\[ g''(-2) = -\frac{f''(-1)}{[f'(-1)]^3} \]
\[ g''(-2) = -\frac{-26}{8^3} = \frac{26}{512} = \frac{13}{256} \]
• We are given $g''(-2) = \frac{a}{b}$ where $a$ and $b$ are positive coprime integers:
\[ a = 13, \quad b = 256 \]
Step 4: Final Answer:
The value of $a$ is 13.
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