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.

  • 1
    $\begingroup$ They forgot a local maximum at $(0,0)$ and that the two local minima are global as well. $\endgroup$ – AlexR Jan 13 '14 at 0:02
  • $\begingroup$ @AlexR OK thanks, but why are my critical points wrong? $\endgroup$ – Emi Matro Jan 13 '14 at 0:03
  • 4
    $\begingroup$ Note ${\sqrt {10}\over 2}={\sqrt 2\cdot\sqrt 5\over 2}={\sqrt5\over\sqrt2}=\sqrt{5\over 2}$. $\endgroup$ – David Mitra Jan 13 '14 at 0:03
  • 1
    $\begingroup$ Multiply and divide by $\sqrt{2}$ $\endgroup$ – IAmNoOne Jan 13 '14 at 0:03
  • 1
    $\begingroup$ @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. $\endgroup$ – 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 $x$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.