FORM GR1268 Question 28

Question:
Let \(f\) be a one-to-one (injective), positive-valued function defined on \(\mathbb R\). Assume that \(f\) is differentiable at \(x=1\) and that in the \(xy\)-plane the line \(y-4=3(x-1)\) is tangent to the graph of \(f\) at \(x=1\). Let \(g\) be the function defined by \(g(x)=\sqrt x\) for \(x\ge0\). which of the following is FALSE?

(A) \(f'(1)=3\)

(B) \(\displaystyle\left(f^{-1}\right)^{\prime}(4)=\frac{1}{3}\)

(C) \((fg)'(1)=5\)

(D) \((g\circ f)'(1)=\frac{1}{2}\)

(E) \((g\circ f)(1)=2\)

Answer:
(D)

Answer Key:
The line \(y-4=3(x-1)\) has a constant derivative, \(3\).
Because \(f\) is tangent to the line at \(x=1\), we have \(f'(1)=3\).

The line goes through the point \((x,y)=(1,4)\).
The slope of \(f^{-1}\) is the reciprocal of the slope of \(f\) at that point.
Thus, (B) is true.

For (C), since \(\displaystyle g'(x)=\frac{1}{2\sqrt x}\), we have
\[\begin{align}
(fg)'(1)&=f'(1)g(1)+f(1)g'(1)\\
&=3\cdot\sqrt1+4\cdot\frac{1}{2\sqrt1}\\
&=5.
\end{align}\]

For (D),
\[\begin{align}
(g\circ f)'(1)&=(g(f(x))'(1)\\
&=\frac{dg}{df}(f(1))\frac{df}{dx}(1)\\
&=\frac{1}{2\sqrt{f(1)}}\cdot3\\
&=\frac{1}{2\sqrt4}\cdot 3\\
&=\frac{3}{4}.
\end{align}\]

For (E), \((g\circ f)(1)=g(f(1))=g(4)=\sqrt{4}=2\).