Help with local extrema of $f(x)=x^4-5x^2$ 
Find the coordinates of any local extreme points and inflection points of the function $f(x)=x^4-5x^2$

My try: 
Find critical points: $f^{\prime}(x)=4x^3-10x=0$
$f^{\prime}(x)=2x(2x^2-5)=0 \implies x=0, x=\pm\sqrt{\dfrac{5}{2}}$ 
I would then use the critical points to determine where the function is increasing/decreasing and by inputing critical point $c$ into $f^{\prime\prime}(c)$, I would determine local min/max.
This is wrong though, because the answers are:
local min: $\left( \dfrac{-\sqrt{10}}{2},\dfrac{-25}{4}  \right)$, $\left( \dfrac{\sqrt{10}}{2},\dfrac{-25}{4}  \right)$
inflection points: $\left( \dfrac{-\sqrt{30}}{6},\dfrac{-125}{36}  \right)$, $\left( \dfrac{\sqrt{30}}{6},\dfrac{-125}{36}  \right)$  
What am I doing wrong and how do I do it correctly? Thanks. 
 A: From where you left off with the critical points you correctly found, we determine their $y$-coordinates as followed
$$\begin{aligned}
f(0) &= 0\\
f\left(-\dfrac{\sqrt{10}}{2} \right) &= -\dfrac{25}{4}\\
f\left(\dfrac{\sqrt{10}}{2} \right) &= -\dfrac{25}{4}
\end{aligned}$$
To determine the nature of those critical points, we use the Second Derivative Test.  First, we find the second derivative of $f(x)$.  Then, we check each critical point.  Here is how one uses the test:
$$\begin{aligned}
f'(x) &= 4x^3 - 10x\\
f''(x) &= 12x^2 - 10\\
f''(0) &= -10 < 0\\
f''\left(-\dfrac{\sqrt{10}}{2} \right) = f''\left(\dfrac{\sqrt{10}}{2} \right) &= 20 > 0
\end{aligned}$$
This shows that the points $\left(\dfrac{\sqrt{10}}{2}, -\dfrac{25}{4}\right)$ and $\left(-\dfrac{\sqrt{10}}{2}, -\dfrac{25}{4} \right)$ are local minima while $(0,0)$ is local maximum.
Finding the inflection points should be extremely easy.  All you need to do is to use $f''(x)$ and set it equal to $0$.  Then, find the values of $x$.
