# Find the extreme points of the function $g(x):=(x^4-2x^2+2)^{1/2}, x∈[-0.5,2]$

I need to find the extreme points of the function $$g(x):=(x^4-2x^2+2)^{1/2}, x∈[-0.5,2]$$ I first found $$f'(x)=\frac{{4x^3-4x}}{2\sqrt {x^4-2x^2+2}}$$ and made $f'(x)=0$ to find all the roots of the function, $x_1=0, x_2=1, x_3=-1$ but since $x_3$ is out of the domain I didn't consider it. Now I have $4$ candidates for the extreme points for this function, namely $x_1, x_2, r_1=-0.5, r_2=2$, where $r_1, r_2$ are the ends of the domain. I then put these candidates back into $f(x)$ and found that $$f(x_2)<f(r_1)<f(x_1)<f(r_2)$$ showing that $x_2$ is the global minimum and $r_2$ is the global maximum.

But I can't seem to figure out the local maximum and local minimum of the function. I tried making a sign table for the function:

But I have no idea how to determine that $x_1$ is the local maximum and $r_1$ is the local minimum.

PS - Sorry for the terrible sign graph, I had to use an online graphing tool.

(...) and found that $$f(x_2)<f(r_1)<f(x_1)<f(r_2)$$ showing that $x_2$ is the global minimum and $r_2$ is the global maximum.

Careful: the (global) extremes are $f(x_2)$ and $f(r_2)$ respectively, they occur in $x_2$ and $r_2$.

But I can't seem to figure out the local maximum and local minimum of the function. I tried making a sign table for the function:

The sign of $f$ doesn't help you. You either make a sign table of its derivative $f'$ or you look at higher order derivates.

Looking at the sign of $f'$, which comes down to the sign of the numerator since the denominator is positive so you only need the sign of $x^3-x=x(x-1)(x+1)$, you'll see that:

• it goes from positive to negative in $x=0$, so $f$ goes from increasing to decreasing and attains a local maximum there;
• it goes from negative to positive in $x=1$, so $f$ goes from decreasing to increasing and attains a local minimum there.

PS - Sorry for the terrible sign graph, I had to use an online graphing tool.

No need; this is a well written and documented question!

• But in my answer booklet it says that $r_1=-0.5$ is the local minimum, since $f$ is a continuous function and there must be a minimum in the interval $[-0.5,0]$ and because in the open interval of $(-0.5,0)$ there are no extreme points, $r_1$ must be a local minimum - which doesn't make much sense to me. And thanks :) – Ski Mask Jan 16 '18 at 16:13
• Yes, there is a local minimum. You can use the sign table of $f'$ to see this since $f'$ is positive for $x>-0.5$ which means $f$ is increasing there - hence it attains a (local) minimum in $-0.5$. – StackTD Jan 16 '18 at 16:18
• but let's say that I want to use the second derivative test, $f''(x)>0$ it's a minimum at that point and if $f''(x)<0$ it's a max, would you still get the same result? – Ski Mask Jan 16 '18 at 21:51
• The second derivative test is useful when the first derivative is 0, which you can use to identify points in the open interval where extremes occur; not on the boundary. Note that the derivative doesn't have to be 0 to have an extreme at one of the end points of the interval. – StackTD Jan 16 '18 at 22:13

I don't know if my answer is suitable, but in this case you can easly avoid derivate. Proceed like this. Since the function $x\mapsto \sqrt{x}$ is strictly increasing, the function $g$ will achieve extreme values when $x^4-2x^2+2$ does.

Since we have $$0\leq x^2 \leq 4$$ we have $$-1\leq x^2-1 \leq 3$$ so $$0\leq (x^2-1)^2 \leq 9$$ and thus $$1\leq (x^2-1)^2+1 \leq 10$$

So $g_{\max} = \sqrt{10}$ when $x=2$ and $g_{\min} = 1$ when $x= 1$.

Completing the square:

$y= [(x^2-1)^2 +1]^{1/2}$, $0.5 \le x \le 2.$

Set $z: =(x^2-1)^2 \ge 0$; then

$0 \le z \le 9.$

Since $f(x)=√x$ an increasing function:

$\min(y) = [ z_{min}+1]^{1/2} = 1$, where $z_{min}=0.$

$\max(y) = [z_{max} +1]^{1/2} = [9+1]^{1/2}$ $= \sqrt{10},$ where $z_{max} =9.$

Note:

$z= (x^2-1)^2$ , where $-0.5 \le x \le 2.$

$z_{min}=0$ for $x^2=1$, i.e $x=1$.

$z_{max} =9$ for $x^2 =4$, i.e. $x=2$.