Maximize the function $f(x) = (1-x) (1+x)^3$ Maximize the function $f(x) = (1-x)(1+x)^3.$
I took the first derivative of $f(x)$ and got $(1+x)^2 (2-4x)$
From there I got $2$ critical points of: $-1, \frac{1}{2}$
Then $I$ took the second derivative of $f(x)$ ($-12x^2 -12x$) and plugged in the critical points to get $0$ and $-6$.
I know I am doing it wrong can anybody help. The right answer is $\frac{27}{16}$.
 A: You're on the right track.
Note that, when $x=\dfrac12, \;f(x)=(1-x)(1+x)^3=\dfrac12\dfrac{27}{8}=\dfrac{27}{16}$,
and $f'(x)=(1+x)^2(2-4x)=0$ and $f''(x)=-12x^2-12x<0$, so that's a maximum.
A: Why do you think you did something wrong? Yes, $-1$ and $\frac12$ are critical points. Besides, $f''\left(\frac12\right)=-9<0$. So, $f$ has a local maximum at $\frac12$. And it has no other local maximum. So, since $\lim_{x\to\pm\infty}f(x)=-\infty$, the maximum is attained at $\frac12$.
A: One of the stationary points has first and second derivative both $0$, so it's a turning point if its third derivative vanishes and the fourth doesn't (which it won't, as $f$ is a quartic). It helps to write $y=1+x$ so $f=y^3(2-y)=2y^3-y^4$, and if $y=3/2$ we get a maximum with $f=27/16$, whereas $y=0$ is a point of inflection with $f=0,\,f^{(3)}=\left.(12-24y)\right|_{y=0}=12\ne0$.
A: Since $f(x) $ is non-negative only when $ |x| \leqslant 1$, then we only need to care when $|x|\leqslant 1$.
With that, we know that $ 1-x$ and $ 1 + x$ are non-negative numbers. Thus, we can apply the AM-GM inequality:
$$ AM\left(1 - x , \frac{1+x}3 , \frac{1+x}3, \frac{1+x}3 \right) \geqslant GM \left (1 - x , \frac{1+x}3 , \frac{1+x}3, \frac{1+x}3 \right) $$
Simplifying the above gives $$\frac12 \geqslant \left ( \frac{(1-x)(1+x)^3}{3^3} \right )^{1/4}  $$
Rearranging gives $ (1-x)(1+x)^3 \leqslant \boxed{\frac{27}{16}}.$
The maximum value is attained when $ 1-x = \dfrac{1+x}3 \Leftrightarrow x = \dfrac12. $
