# 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.

• They forgot a local maximum at $(0,0)$ and that the two local minima are global as well. – AlexR Jan 13 '14 at 0:02
• @AlexR OK thanks, but why are my critical points wrong? – Emi Matro Jan 13 '14 at 0:03
• Note ${\sqrt {10}\over 2}={\sqrt 2\cdot\sqrt 5\over 2}={\sqrt5\over\sqrt2}=\sqrt{5\over 2}$. – David Mitra Jan 13 '14 at 0:03
• Multiply and divide by $\sqrt{2}$ – IAmNoOne Jan 13 '14 at 0:03
• @user436158 They are not. $$\frac{\sqrt{10}}2 = \sqrt{\frac{10}4} = \sqrt{\frac52}$$ But usually you try to eliminate square roots in the denominator, so $\frac{\sqrt{10}}2$ is chosen as a "standard" representation. – AlexR Jan 13 '14 at 0:05

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 $x$.

• Thanks. How would I show that the two local minima are also global minima? – Emi Matro Jan 13 '14 at 0:34