Question

if $f(-3)=4$, and $f(-3)=-\frac{1}{2}$. let $f^{-1}(x)$ be the inverse function of $f(x)$. $(f^{-1})(4)=$
write the equation of the tangent line to $f^{-1}(x)$ at $x = 4$. $y=$

Explanation:

Step1: Recall inverse - function derivative formula

The formula for the derivative of the inverse function is \((f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}\). Given \(a = 4\), and since \(f(-3)=4\), then \(f^{-1}(4)=-3\).

Step2: Calculate \((f^{-1})'(4)\)

Substitute \(a = 4\) into the formula. We know that \(f^{-1}(4)=-3\) and \(f'(-3)=-\frac{1}{2}\). So \((f^{-1})'(4)=\frac{1}{f'(-3)}=\frac{1}{-\frac{1}{2}}=-2\).

Step3: Find a point on \(y = f^{-1}(x)\)

Since \(f(-3)=4\), then the point on the inverse - function \(y = f^{-1}(x)\) is \((4,-3)\) (because if \((x_0,y_0)\) is on \(y = f(x)\), then \((y_0,x_0)\) is on \(y = f^{-1}(x)\)).

Step4: Write the equation of the tangent line

The point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(4,-3)\) and \(m=(f^{-1})'(4)=-2\).
\[y-(-3)=-2(x - 4)\]
\[y + 3=-2x+8\]
\[y=-2x + 5\]

Answer:

\((f^{-1})'(4)=-2\)
\(y=-2x + 5\)