Question

question consider the function f(x) below. over what interval(s) is the function concave up? give your answer in interval notation. note that if the function meets the given criteria over multiple intervals, utilize a union symbol between them. (e.g. (0,1) ∪ (2,3)) f(x)=3x^4 + 4x^3+2x + 6 provide your answer below: f(x) is concave up over the interval(s)

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:

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