Question

1. let ( f ) be a differentiable function such that ( f(3)=15, f(6)=3, f^{prime}(3)=-8 ), and ( f^{prime}(6)=-2 ). the function ( g ) is differentiable and ( g(x)=f^{-1}(x) ) for all ( x ). what is the value of ( g^{prime}(3) )?

a ( \frac{1}{2} )
b ( \frac{1}{8} )
c ( \frac{1}{6} )
d ( \frac{1}{3} )
e the value of ( g^{prime}(3) ) cannot be determined from the information given.

Explanation:

Step1: Recall the inverse function derivative formula

The formula for the derivative of an inverse function is \((f^{-1})'(y)=\frac{1}{f'(x)}\), where \(y = f(x)\) (i.e., \(x = f^{-1}(y)\)). Here, \(g(x)=f^{-1}(x)\), so we need to find \(g'(3)\), which is \((f^{-1})'(3)\).

Step2: Find \(x\) such that \(f(x)=3\)

We know that \(f(6) = 3\), so when \(y = 3\), \(x = 6\) (since \(f(6)=3\) implies \(f^{-1}(3)=6\)).

Step3: Apply the inverse function derivative formula

Using the formula \((f^{-1})'(y)=\frac{1}{f'(x)}\) with \(y = 3\) and \(x = 6\) (because \(f(6)=3\)), we have \((f^{-1})'(3)=\frac{1}{f'(6)}\).

Step4: Substitute the value of \(f'(6)\)

We are given that \(f'(6)=-2\), so \((f^{-1})'(3)=\frac{1}{-2}=-\frac{1}{2}\)? Wait, no, wait. Wait, no, let's check again. Wait, the formula is \((f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}\). So we need to find \(f^{-1}(3)\) first. Since \(f(6) = 3\), then \(f^{-1}(3)=6\). Then \((f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(6)}\). Since \(f'(6)=-2\), then \((f^{-1})'(3)=\frac{1}{-2}\)? Wait, no, that can't be. Wait, no, the formula is \((f^{-1})'(y)=\frac{1}{f'(x)}\) where \(y = f(x)\). So if \(y = 3\) and \(x = 6\) (because \(f(6)=3\)), then \((f^{-1})'(3)=\frac{1}{f'(6)}\). But \(f'(6)=-2\), so \((f^{-1})'(3)=\frac{1}{-2}=-\frac{1}{2}\)? Wait, but let's check the options. Option A is \(-\frac{1}{2}\)? Wait, no, wait the options are A: \(-\frac{1}{2}\), B: \(-\frac{1}{8}\), C: \(\frac{1}{6}\), D: \(\frac{1}{3}\), E: cannot be determined. Wait, but wait, did I make a mistake? Wait, \(f(3)=15\), \(f(6)=3\), \(f'(3)=-8\), \(f'(6)=-2\). So \(g(x)=f^{-1}(x)\), so \(g'(3)=(f^{-1})'(3)\). By the inverse function derivative rule, \((f^{-1})'(y)=\frac{1}{f'(x)}\) where \(f(x)=y\). So when \(y = 3\), \(x = 6\) (since \(f(6)=3\)), so \((f^{-1})'(3)=\frac{1}{f'(6)}=\frac{1}{-2}=-\frac{1}{2}\). But wait, the option A is \(-\frac{1}{2}\). Wait, but let's check again. Wait, maybe I mixed up. Wait, no, the formula is correct. So \(g'(3)=(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}\). Since \(f^{-1}(3)=6\) (because \(f(6)=3\)), then \(f'(f^{-1}(3))=f'(6)=-2\), so \((f^{-1})'(3)=\frac{1}{-2}=-\frac{1}{2}\). So the answer should be A.

Answer:

A. \(-\frac{1}{2}\)