Functions
Inverse functions and higher-order derivatives
MJAT_TS1_P2
Grade 12
Question:
Suppose $f(x)$ is a differentiable invertible function with $f'(x) \neq 0$ and $h(x) = \displaystyle\int_1^x f(t)\,dt$. Given that $f(1) = f'(1) = 1$ and $g(x)$ is the inverse of $f(x)$. Let $G(x) = x^2 g(x) - x\,h(g(x))$ for all $x \in \mathbb{R}$. Then $G''(1) =$
Step-by-Step Solution
Key Concept: Since $g = f^{-1}$: $f(g(x)) = x \Rightarrow f'(g(x))g'(x) = 1$. At $x=1$: $f'(g(1))g'(1) = 1$. Since $g(1) = f^{-1}(1) = 1$ (as $f(1)=1$), $f'(1)g'(1)=1 \Rightarrow g'(1)=1$. Also $h'(x) = f(x)$ and $h(g(1)) = h(1) = 0$.
$G'(x) = 2xg(x) + x^2g'(x) - h(g(x)) - x\cdot f(g(x))\cdot g'(x) = 2xg(x) + x^2g'(x) - h(g(x)) - x^2g'(x) = 2xg(x) - h(g(x))$. $G''(x) = 2g(x) + 2xg'(x) - f(g(x))g'(x) = 2g(x) + xg'(x)$. At $x=1$: $G''(1) = 2(1) + 1(1) = \mathbf{3}$.
Correct Answer: 3