Question

Determine the value of \({G^\prime }(2)\).

Short answer

The value of \({G^\prime }(2)\) is \( - \frac{1}{9}\).

Step-by-step solution

Given

The function\(f\)is differentiable and its inverse exists.

Also \(f(3) = 2,{f^\prime }(3) = \frac{1}{9}\) and \(G(x) = \frac{1}{{{f^{ - 1}}(x)}}\).

Concept of Inverse function

If\(f\)is a one-to-one differentiable function with inverse function\({f^{ - 1}}\)and\({f^\prime }\left( {{f^{ - 1}}(a)} \right) \ne 0\), then the inverse function is differentiable at\(a\)and\({\left( {{f^{ - 1}}} \right)^\prime }(a) = \frac{1}{{{f^\prime }\left( {{f^{ - 1}}(a)} \right)}}\).

Use the theorem to find the value of \({G^\prime }(2)\)

From the given information,\(f(3) = 2\).

Thus,\({f^{ - 1}}(2) = 3\).

\(\begin{array}{c}G(2) = \frac{1}{{{f^{ - 1}}(2)}}\\G(2) = \frac{1}{3}\end{array}\)

With the help of the theorem If\(f\)is a one-to-one differentiable function with inverse function\({f^{ - 1}}\)and\({f^\prime }\left( {{f^{ - 1}}(a)} \right) \ne 0\), then the inverse function is differentiable at\(a\)and\({\left( {{f^{ - 1}}} \right)^\prime }(a) = \frac{1}{{{f^\prime }\left( {{f^{ - 1}}(a)} \right)}}\)obtain the value of\({\left( {{f^{ - 1}}} \right)^\prime }(2)\)as follows:

\(\begin{array}{c}{\left( {{f^{ - 1}}} \right)^\prime }(2) = \frac{1}{{{f^\prime }\left( {{f^{ - 1}}(2)} \right)}}\\ = \frac{1}{{{f^\prime }(3)}}\\ = \frac{1}{{\left( {\frac{1}{9}} \right)}}\\ = 9\end{array}\)

Since,\(G(x) = \frac{1}{{{f^{ - 1}}(x)}}\), differentiate\(G(x) = \frac{1}{{{f^{ - 1}}(x)}}\)with respect\(x\).

\(\begin{array}{c}{G^\prime }(x) = \frac{d}{{dx}}\left( {\frac{1}{{{f^{ - 1}}(x)}}} \right)\\ = \frac{{ - 1}}{{{f^\prime }\left( {{f^{ - 1}}(x)} \right)}}\end{array}\)

Now substitute\(x = 2\)in\({G^\prime }(x) = \frac{{ - 1}}{{{f^\prime }\left( {{f^{ - 1}}(x)} \right)}}\).

\(\begin{array}{c}{G^\prime }(2) = \frac{{ - 1}}{{{f^\prime }\left( {{f^{ - 1}}(2)} \right)}}\\ = \frac{{ - 1}}{{{{\left( {{f^{ - 1}}} \right)}^\prime }(2)}}\\ = \frac{{ - 1}}{9}\end{array}\)

Thus, the value of \({G^\prime }(2)\) is \(\frac{{ - 1}}{9}\).