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 $(x^n)'=nx^{n - 1}$, for $f(x)=x^{4}+4x^{3}+6x^{2}+4x$, 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 inflection points

Set $f''(x)=0$. Since $12(x + 1)^{2}=0$, then $x=-1$. To find the $y$ - value, substitute $x = - 1$ into $f(x)$: $f(-1)=(-1)^{4}+4(-1)^{3}+6(-1)^{2}+4(-1)=1-4 + 6-4=-1$. So the inflection point is $(-1,-1)$.

Step5: Determine concavity

Since $f''(x)=12(x + 1)^{2}\geq0$ for all real $x$, the function is concave up on $(-\infty,\infty)$ and concave down on NONE.

Answer:

Concave up on: $(-\infty,\infty)$
Concave down on: NONE
Inflection points $(x,y)$: $(-1,-1)$