if $f(-3)=4$, and $f(-3)=-\frac{1}{2}$. let $f^{-1}(x)$ be the inverse function of $f(x)$…

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

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

Answer

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)