Question

Find the critical numbers of the function.

24. \[f(x) = {x^3} + 6{x^2} - 15x\].

Short answer

The critical point is \(x = - 5\) and \(x = 1\).

Step-by-step solution

Given data

The given function is \(f(x) = {x^3} + 6{x^2} - 15x\).

Concept of Differentiation

Differentiation is a method of finding the derivative of a function. Differentiation is a process, where we find the instantaneous rate of change in function based on one of its variables.

Determine the derivative of the function

Obtain the first derivative of the given function \(f(x) = {x^3} + 6{x^2} - 15x\).

\(\begin{aligned}{l}{f^\prime }(x) &= \frac{d}{{dx}}\left( {{x^3} + 6{x^2} - 15x} \right)\\{f^\prime }(x)\; &= \frac{d}{{dx}}\left( {{x^3}} \right) + \frac{d}{{dx}}\left( {6{x^2}} \right) - \frac{d}{{dx}}(15x)\\{f^\prime }(x) &= 3 \cdot \left( {{x^{3 - 1}}} \right) + 6 \cdot \left( {2{x^{2 - 1}}} \right) - 15\end{aligned}\) ………. (1)

Apply power rule as in the equation 1: \(3{x^2} + 12x - 15\)

Take \({f^\prime }(x) = 0\) to obtain the critical point.

\(\begin{aligned}{c}3{x^2} + 12x - 15 &= 0\\3\left( {{x^2} + 4x - 5} \right) &= 0\\{x^2} + 4x - 5 &= 0\\(x + 5)(x - 1) &= 0\end{aligned}\)

Thus, the critical number of \(f(x) = {x^3} + 6{x^2} - 15x\) is, \(x = - 5\) and \(x = 1\).