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Q26E

Expert-verifiedFound in: Page 209

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Find the critical numbers of the function.**

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

There is no critical point for the function.

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

**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.**

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, \(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\) 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.

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