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Let $f(x)= 10xe^x$

$(a)$ Find the exact value of $x$ so that $f ''(x) = 0$.

I tried:

$$\begin{align}f'(x)& = 10e^x\\f''(x)&= e^x\end{align}$$ but at that point, the $f''(x)$ would never be a zero. So what is my mistake?

$(b)$ For what interval is $f(x)$ concave up?

I wonder how to know the concavity after knowing the equation

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  • $\begingroup$ One mistake is not using the product rule to evaluate $(10xe^x)'$. $\endgroup$ – David Mitra Jun 18 '14 at 23:45
  • $\begingroup$ Once you have the correct second derivative function, $ \ f(x) \ $ is "concave upward" wherever $ \ f \ ''(x) \ > \ 0 \ $ and "concave downward" wherever $ \ f \ ''(x) \ < \ 0 \ $ . Keep in mind that the exponential factor $ \ e^x \ $ is always positive. $\endgroup$ – colormegone Jun 18 '14 at 23:49
  • $\begingroup$ @DavidMitra thanks for the reminder $\endgroup$ – John Jun 18 '14 at 23:50
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(a) $f'(x)=10e^x + 10xe^x$.

$f''(x)=10e^x + 10e^x + 10xe^x = 20e^x + 10xe^x =10e^x(2+x)$

So, at $x=-2$ you have $f''(x)=0$.

(b) Hint: If $f''(x)>0$, $f$ is concave up at $x$, and if $f''(x)<0$, $f$ is concave down at $x$.

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  • $\begingroup$ Thaanks !What did you do to go from f'(x) to f''(x)? $\endgroup$ – John Jun 18 '14 at 23:49
  • $\begingroup$ I computed it using the product rule. $\endgroup$ – Twink Jun 18 '14 at 23:50
  • $\begingroup$ @John, derivation using the product rule. $f''(x) = (10e^x + 10xe^x)' = 10 (e^x)' + 10(x)'e^x + 10x(e^x)' = 10e^x + 10 e^x + 10xe^x = 10(2+x)e^x$ $\endgroup$ – Graham Kemp Jun 18 '14 at 23:52
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$(a)$ Since $f(x)=10x\cdot e^x$ we will use the product rule to obtain $f'(x)$ and $f''(x)$. So $f'(x)=10x\cdot e^x+10e^x$ and $f''(x)=10x\cdot e^x+20e^x$. Set $f''(x)=0$. So $10x\cdot e^x+20e^x=0$ implies that $10e^x(x+2)=0$. We know that $10e^x$ is never $0$ and so $x=-2$ will make $f''(x)=0$.

$(b)$ If $x<-2$, then the function is concave down. If $x>-2$, then the function is concave up.

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