Question

question 3
consider the function $f(x) = x^2 e^{6x}$.
$f(x)$ 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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Explanation:

Step1: Find first derivative

Use product rule: $(uv)'=u'v+uv'$, where $u=x^2$, $v=e^{6x}$.
$f'(x) = 2x e^{6x} + x^2 \cdot 6e^{6x} = e^{6x}(6x^2 + 2x)$

Step2: Find second derivative

Differentiate $f'(x)$ with product rule.
$f''(x) = 6e^{6x}(6x^2 + 2x) + e^{6x}(12x + 2) = e^{6x}(36x^2 + 12x + 12x + 2) = e^{6x}(36x^2 + 24x + 2)$

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

Since $e^{6x}
eq 0$, solve $36x^2 +24x +2=0$. Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ with $a=36, b=24, c=2$.

$$\begin{align*} x&=\frac{-24\pm\sqrt{24^2-4(36)(2)}}{2(36)}\\ &=\frac{-24\pm\sqrt{576-288}}{72}\\ &=\frac{-24\pm\sqrt{288}}{72}\\ &=\frac{-24\pm12\sqrt{2}}{72}\\ &=\frac{-2\pm\sqrt{2}}{6} \end{align*}$$

Step4: Classify concavity intervals

Test values in intervals:

• For $(-\infty, \frac{-2-\sqrt{2}}{6})$: pick $x=-1$, $f''(-1)=e^{-6}(36-24+2)>0$ → concave up
• For $(\frac{-2-\sqrt{2}}{6}, \frac{-2+\sqrt{2}}{6})$: pick $x=0$, $f''(0)=e^{0}(0+0+2)>0$? No, $f''(0)=2$? Wait, no: $36(0)^2+24(0)+2=2>0$? Correction: pick $x=-0.3$, $36(0.09)+24(-0.3)+2=3.24-7.2+2=-1.96<0$ → concave down
• For $(\frac{-2+\sqrt{2}}{6}, \infty)$: pick $x=0$, $f''(0)=2>0$ → concave up

Answer:

$C = \frac{-2-\sqrt{2}}{6}$, $D = \frac{-2+\sqrt{2}}{6}$

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