Question

Let $f(x) = x^5 + 2x^3 + 3x + 1$, $x \in \mathbb{R}$, and $g(x)$ be a function such that $g(f(x)) = x$ for all $x \in \mathbb{R}$. Then $\frac{g(7)}{g'(7)}$ is equal to:

• A. 7
• B. 42
• C. 1
• D. 14

Answer 1

Answer: (D)

14

Answer 2

Given:

$f(x) = x^5 + 2x^3 + 3x + 1$

Then,

$f'(x) = 5x^4 + 6x^2 + 3$

Calculate $f'(1)$:

$f'(1) = 5 \cdot 1^4 + 6 \cdot 1^2 + 3 = 14$

Since $g(f(x)) = x$, by differentiation, we get:

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

For $f(x) = 7$:

$x^5 + 2x^3 + 3x + 1 = 7$

This implies $x = 1$, so $f(1) = 7$.

Then $g(7) = 1$.

Now,

$g'(7)f'(1) = 1 \Rightarrow g'(7) = \frac{1}{f'(1)} = \frac{1}{14}$

Thus, $\frac{g(7)}{g'(7)} = \frac{1}{\frac{1}{14}} = 14$