Question

find the intervals of concavity and inflection points. list all answers in each category separated by commas, and enter none if none exist.

$f(x)=x^{4}+4x^{3}+6x^{2}+4x$

concave up on:

concave down on:

inflection points $(x,y):$

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=4x^{3}+12x^{2}+12x + 4$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=12x^{2}+24x + 12$.

Step3: Factor the second - derivative

$f''(x)=12(x^{2}+2x + 1)=12(x + 1)^{2}$.

Step4: Find the intervals of concavity

Set $f''(x)=0$. Since $12(x + 1)^{2}=0$, we get $x=-1$.
Test intervals:
Choose a test point in the interval $(-\infty,-1)$, say $x=-2$. Then $f''(-2)=12(-2 + 1)^{2}=12>0$.
Choose a test point in the interval $(-1,\infty)$, say $x = 0$. Then $f''(0)=12(0 + 1)^{2}=12>0$.
Since $f''(x)\geq0$ for all $x\in(-\infty,\infty)$ (equality holds only at $x=-1$), the function is concave up on $(-\infty,\infty)$ and concave down on NONE.

Step5: Find the inflection points

Since the concavity does not change ( $f''(x)$ does not change sign), the inflection points are NONE.

Answer:

Concave up on: $(-\infty,\infty)$
Concave down on: NONE
Inflection points $(x,y)$: NONE