Question

consider the function $f(x)=x^{2}e^{3x}$.
$f(x)$ has two inflection points at $x = c$ and $x = d$ with $c < d$
where $c$ is
and $d$ is
finally for each of the following intervals, tell whether $f(x)$ is concave up or concave down.
$(-infty,c)$: select an answer
$(c,d)$: select an answer
$(d,infty)$ select an answer
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question 4
which graph of $f(x)$ satisfies
$f(x)>0$ on $(-infty,-1) cup (1,infty)$ and
$f(x)<0$ on $(-1,1)$

Explanation:

Step1: Find first derivative

Use product rule: $(uv)'=u'v+uv'$, let $u=x^2$, $v=e^{3x}$.
$u'=2x$, $v'=3e^{3x}$.
$f'(x)=2x e^{3x} + 3x^2 e^{3x} = e^{3x}(3x^2+2x)$

Step2: Find second derivative

Differentiate $f'(x)$ with product rule: let $u=3x^2+2x$, $v=e^{3x}$.
$u'=6x+2$, $v'=3e^{3x}$.
$f''(x)=(6x+2)e^{3x} + 3(3x^2+2x)e^{3x} = e^{3x}(9x^2+12x+2)$

Step3: Solve $f''(x)=0$

Since $e^{3x}>0$ for all $x$, solve $9x^2+12x+2=0$.
Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=9$, $b=12$, $c=2$.
$x=\frac{-12\pm\sqrt{144-72}}{18}=\frac{-12\pm\sqrt{72}}{18}=\frac{-12\pm6\sqrt{2}}{18}=\frac{-2\pm\sqrt{2}}{3}$

Step4: Identify inflection points

Order the roots: $C=\frac{-2-\sqrt{2}}{3}\approx-1.138$, $D=\frac{-2+\sqrt{2}}{3}\approx-0.195$

Step5: Test concavity intervals

• For $(-\infty,C)$: pick $x=-2$, $f''(-2)=e^{-6}(36-24+2)>0$, concave up.
• For $(C,D)$: pick $x=-0.5$, $f''(-0.5)=e^{-1.5}(2.25-6+2)<0$, concave down.
• For $(D,\infty)$: pick $x=0$, $f''(0)=e^{0}(0+0+2)>0$, concave up.

A function is concave up when $f''(x)>0$ (curves upward) and concave down when $f''(x)<0$ (curves downward). The graph will curve upward on $(-\infty,-1)\cup(1,\infty)$ and curve downward on $(-1,1)$, with inflection points at $x=-1$ and $x=1$.

Answer:

$C=\boldsymbol{\frac{-2-\sqrt{2}}{3}}$, $D=\boldsymbol{\frac{-2+\sqrt{2}}{3}}$

• $(-\infty, C)$: Concave Up
• $(C, D)$: Concave Down
• $(D, \infty)$: Concave Up

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For the second part (Question 4):