find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer…

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)\n\n$f(x)=2x^{3}+x^{2}+2x$\n\n$x =$\n\nresources\nread it

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)\n\n$f(x)=2x^{3}+x^{2}+2x$\n\n$x =$\n\nresources\nread it

Answer

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). [ \begin{align*} f^\prime(x)&=2\times3x^{2}+2x + 2\ &=6x^{2}+2x + 2 \end{align*} ]

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}). [ \begin{align*} x&=\frac{-2\pm\sqrt{2^{2}-4\times6\times2}}{2\times6}\ &=\frac{-2\pm\sqrt{4 - 48}}{12}\ &=\frac{-2\pm\sqrt{- 44}}{12} \end{align*} ] 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