Question

Find the critical numbers of the function.

26. \(f(x) = 2{x^3} + {x^2} + 2x\).

Short answer

There is no critical point for the function.

Step-by-step solution

Given data

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

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 \(f(x) = 2{x^3} + {x^2} + 2x\).

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

Apply power rule as in the equation: \(6{x^2} + 2x + 2\)

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

\(\begin{array}{c}6{x^2} + 2x + 2 = 0\\2\left( {3{x^2} + x + 1} \right) = 0\\3{x^2} + x + 1 = 0\end{array}\)

Apply quadratic formula and simplify the expression

Apply quadratic formula, \(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\) and obtain the roots.

\(\begin{aligned}{l}x &= \frac{{ - 1 \pm \sqrt {1 - 12} }}{6}\\x &= \frac{{ - 1 \pm \sqrt { - 11} }}{6}\end{aligned}\)

Thus, the roots are, \(x = \frac{{ - 1 + \sqrt { - 11} }}{6},x = \frac{{ - 1 - \sqrt { - 11} }}{6}\).

Here, the roots are not real as the discriminant \(( - 11)\) is less than 0.

Therefore, it fails to satisfy the conditions of definition of the critical numbers. Hence there is no critical point for the given function. Thus, there is no critical point for the function.