Question

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)

$f(x)=2x^{3}+x^{2}+2x$

$x =$

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Explanation:

Step1: Differentiate the function

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(y = f(x)=2x^{3}+x^{2}+2x\), we have \(f^\prime(x)=(2x^{3})^\prime+(x^{2})^\prime+(2x)^\prime\).

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Step2: Find the roots of the derivative

Set \(f^\prime(x) = 0\), so \(6x^{2}+2x + 2=0\). The quadratic formula for \(ax^{2}+bx + c = 0\) (\(a = 6\), \(b = 2\), \(c = 2\)) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

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Since the discriminant \(\Delta=b^{2}-4ac=4-48=-44<0\), there are no real - valued solutions for \(f^\prime(x) = 0\). And \(f^\prime(x)=6x^{2}+2x + 2\) is a polynomial, so it is defined for all real \(x\).

Answer:

DNE