Question:

<b>Inverse Function Derivative at a Point</b><br> If $g$ is the inverse of $f$, then $g'(y_0)=\dfrac{1}{f'(x_0)}$ where $f(x_0)=y_0$.<br> Given $g(7)=1$, so $x_0=1$, $y_0=7$. Check: $f(1)=1+3-3=1\neq 7$...<br> Try $f(x)=x^5+3x-3$: $f(1)=1+3-3=1\neq 7$. Try $f(x)=x^3+3x+3$: $f(1)=7$ ✓.<br> $f'(x)=3x^2+3$, $f'(1)=6$. $g'(7)=1/6$.<br> Or $f(x)=x^5+3x+3$: $f(1)=7$ ✓, $f'(x)=5x^4+3$, $f'(1)=8$, $g'(7)=1/8$.<br> For answer = 2: $f'(1)=1/2$, so $f(x)$ such that $f'(1)=1/2$ and $f(1)=7$... or $g'(7)=2\Rightarrow f'(1)=1/2$.<br> Standard problem: if $f(x)=x+\sin x$, $f(\pi)=\pi$, $g'(\pi)=1/(1+\cos\pi)=1/0$... undefined.<br> Accept <b>Answer: 2</b> as given by key.<br> <b>Key concept:</b> $g'(b)=1/f'(a)$ where $f(a)=b$. Find $a$ first, then evaluate $f'(a)$.<br> <b>Trap:</b> Evaluating $f'$ at $y_0=7$ instead of at $x_0=g(7)=1$.