1. **State the problem:** We have a function $f$ with an inverse $f^{-1}$ defined on an interval $I$. Given $f(3) = -4$ and the derivative of the inverse at $-4$ is $(f^{-1})'(-4) = 7$, we need to find $f'(3)$ and then find the equation of the tangent line to the curve of $f^{-1}$ at the point where the $x$-coordinate is $-4$. 2. **Recall the formula for the derivative of the inverse function:** The derivative of the inverse function at a point $y = f(x)$ is given by: $$ (f^{-1})'(y) = \frac{1}{f'(x)} $$ where $y = f(x)$. 3. **Use the given values:** We know $f(3) = -4$, so $x=3$ corresponds to $y=-4$. Also, $(f^{-1})'(-4) = 7$. 4. **Calculate $f'(3)$:** Using the formula: $$ (f^{-1})'(-4) = \frac{1}{f'(3)} $$ Substitute the known value: $$ 7 = \frac{1}{f'(3)} $$ Solve for $f'(3)$: $$ f'(3) = \frac{1}{7} $$ 5. **Find the equation of the tangent line to $f^{-1}$ at $x = -4$:** The point on $f^{-1}$ is $(-4, 3)$ because $f(3) = -4$ means $f^{-1}(-4) = 3$. The slope of the tangent line is $(f^{-1})'(-4) = 7$. Using point-slope form: $$ y - y_1 = m(x - x_1) $$ where $(x_1, y_1) = (-4, 3)$ and $m=7$: $$ y - 3 = 7(x + 4) $$ Simplify: $$ y = 7x + 28 + 3 $$ $$ y = 7x + 31 $$ **Final answers:** - $f'(3) = \frac{1}{7}$ - Equation of tangent line to $f^{-1}$ at $x = -4$ is $y = 7x + 31$