# Calculus / find the value of $x$ so that $f ''(x)=0$

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

• One mistake is not using the product rule to evaluate $(10xe^x)'$. – David Mitra Jun 18 '14 at 23:45
• 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. – colormegone Jun 18 '14 at 23:49
• @DavidMitra thanks for the reminder – John Jun 18 '14 at 23:50

(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 $x$.

• Thaanks !What did you do to go from f'(x) to f''(x)? – John Jun 18 '14 at 23:49
• I computed it using the product rule. – Twink Jun 18 '14 at 23:50
• @John, derivation using the product rule. $f''(x) = (10e^x + 10xe^x)' = 10 (e^x)' + 10(x)'e^x + 10x(e^x)' = 10e^x + 10 e^x + 10xe^x = 10(2+x)e^x$ – Graham Kemp Jun 18 '14 at 23:52

$(a)$ 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.