question consider the function f(x) below. over what interval(s) is the function concave up? give your…

Explanation:

Step1: Find the first - derivative

Differentiate $f(x)=3x^{4}+4x^{3}+2x + 6$ using the power rule. The power rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$. $f^\prime(x)=12x^{3}+12x^{2}+2$

Step2: Find the second - derivative

Differentiate $f^\prime(x)$ using the power rule again. $f^{\prime\prime}(x)=36x^{2}+24x$

Step3: Set the second - derivative greater than 0

We want to find where $f^{\prime\prime}(x)>0$ since a function is concave up when its second - derivative is positive. $36x^{2}+24x>0$ Factor out $12x$: $12x(3x + 2)>0$

Step4: Solve the inequality

Set $12x(3x + 2)=0$, the roots are $x = 0$ and $x=-\frac{2}{3}$. We consider the intervals $(-\infty,-\frac{2}{3})$, $(-\frac{2}{3},0)$ and $(0,\infty)$. For $x\in(-\infty,-\frac{2}{3})$, let $x=-1$, then $12(-1)[3(-1)+2]=12>0$. For $x\in(-\frac{2}{3},0)$, let $x =-\frac{1}{3}$, then $12(-\frac{1}{3})[3(-\frac{1}{3})+2]=-4<0$. For $x\in(0,\infty)$, let $x = 1$, then $12(1)[3(1)+2]=60>0$.

Answer:

$(-\infty,-\frac{2}{3})\cup(0,\infty)$