Question

If $f(x)= x^5 + 2x^3 + 3x + 1$ and $g(f(x)) = x,$ then $\frac{g(1)}{g’(1)}$ is equal to

Answer 1

Correct Answer: 0

To solve for $\frac{g(1)}{g'(1)}$ given $f(x) = x^5 + 2x^3 + 3x + 1$ and $g(f(x)) = x$, we recognize that $g$ is the inverse function of $f$. Thus, $g(f(x)) = x$ implies $f(g(y)) = y$. Differentiating both sides of $g(f(x)) = x$ with respect to $x$, we apply the chain rule: $g'(f(x)) \cdot f'(x) = 1$. Solving for $g'(f(x))$, we find $g'(f(x)) = \frac{1}{f'(x)}$.

Given $f(x) = x^5 + 2x^3 + 3x + 1$, compute $f'(x)$: $f'(x) = 5x^4 + 6x^2 + 3$. To find $g'(1)$, note $f(g(1)) = 1$, so $f(g(1)) = g(1)$. We first compute $g(1)$. Since $f(x)$ is directly one-to-one over real numbers, we need to find $x$ such that $f(x) = 1$. Solving $x^5 + 2x^3 + 3x + 1 = 1$, we get $x^5 + 2x^3 + 3x = 0$, factoring $x$ yields $x(x^4 + 2x^2 + 3) = 0$. Thus $x = 0$; verify $f(0) = 1$. Hence, $g(1) = 0$.

To find $g'(1)$, compute $f'(0) = 5(0)^4 + 6(0)^2 + 3 = 3$. So, $g'(1) = \frac{1}{f'(0)} = \frac{1}{3}$.

Finally, $\frac{g(1)}{g'(1)} = \frac{0}{\frac{1}{3}} = 0$, which falls within the expected range 0,0. Thus, $\frac{g(1)}{g'(1)} = 0$, verified to fit the provided range.