Question

Determine the value of\({\left( {{f^{ - 1}}} \right)^\prime }(4)\), where\(f(x) = 2{x^3} + 3{x^2} + 7x + 4\).

Short answer

The value of \({\left( {{f^{ - 1}}} \right)^\prime }(4) = \frac{1}{7}\).

Step-by-step solution

Given equation

The given equation is \(f(x) = 2{x^3} + 3{x^2} + 7x + 4\).

Definition of Horizontal Line Test

In Horizontal line test a function is said to be one-to one if and only if no horizontal line intersects its graph more than once.

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\({\left( {{f^{ - 1}}} \right)^\prime }(a) = \frac{1}{{{f^\prime }\left( {{f^{ - 1}}(a)} \right)}}\).

Plot the Graph

The given function is observed to be differential.

Use horizontal line test to determine whether the function\(f(x) = 2{x^3} + 3{x^2} + 7x + 4\)is one-to-one.

Sketch the graph of the function \(f(x) = 2{x^3} + 3{x^2} + 7x + 4\) and perform the horizontal line test as shown below in Figure 1.

From Figure 1, it is observed that the horizontal line intersects the curve at one point which means it passes the horizontal line test.

Thus, the given function is one-to-one.

From Figure 1,\(f\left( 0 \right) = 4\).

Thus, \({f^{ - 1}}(4) = 0\)

Use the theorem for calculation

By the theorem, for\(f(x) = 2{x^3} + 3{x^2} + 7x + 4\).

\(\begin{array}{c}{\left( {{f^{ - 1}}} \right)^\prime }(4) = \frac{1}{{{f^\prime }\left( {{f^{ - 1}}(4)} \right)}}\\ = \frac{1}{{{f^\prime }(0)}}\\ = \frac{1}{{6{{(0)}^2} + 6(0) + 7}}\\ = \frac{1}{7}\end{array}\)

Thus, the value of \({\left( {{f^{ - 1}}} \right)^\prime }(4) = \frac{1}{7}\).