1. ## max/min

Find the values of x that give relative extrema for the function $\displaystyle f(x) = 3x^5 - 5x^3$

What is relative maximum and relative minimum of it?

2. Originally Posted by Samantha
Find the values of x that give relative extrema for the function $\displaystyle f(x) = 3x^5 - 5x^3$

What is relative maximum and relative minimum of it?
I might not be right because I'm not sure why it says RELATIVE maximum and minimum, but to get the maximum and minimum of a function, differentiate it and set it equal to 0.

$\displaystyle f(x) = 3x^5 - 5x^3$

$\displaystyle \frac{dy}{dx} \ f(x) = 15x^4 - 15x^2 = 0$

$\displaystyle 0 = 15x^2(x^2 - 1)$

So $\displaystyle x = 1 \ or \ x = -1 \ or \ x = 0$

3. Originally Posted by Samantha
Find the values of x that give relative extrema for the function $\displaystyle f(x) = 3x^5 - 5x^3$

What is relative maximum and relative minimum of it?
f'(x)=15x^2(x+1)(x-1)=0 gives the stationary points x=0,-1,1.

4. Originally Posted by janvdl
I might not be right because I'm not sure why it says RELATIVE maximum and minimum, but to get the maximum and minimum of a function, differentiate it and set it equal to 0.

$\displaystyle f(x) = 3x^5 - 5x^3$

$\displaystyle \frac{dy}{dx} \ f(x) = 15x^4 - 15x^2 = 0$

$\displaystyle 0 = 15x^2(x^2 - 1)$

So $\displaystyle x = 1 \ or \ x = -1 \ or \ x = 0$
I don't understand =(

Is it relative maximum: x=-1, Relative minimum: x=1?

5. Originally Posted by Samantha
I don't understand =(

Is it relative maximum: x=-1, Relative minimum: x=1?
Set these values back into the original equation and see which is the largest and smallest

6. Originally Posted by Samantha
I don't understand =(

Is it relative maximum: x=-1, Relative minimum: x=1?
Consider the sign of its derivative at both sides of the stationary points to determine if they are relative max (min) points.

7. Originally Posted by janvdl
Set these values back into the original equation and see which is the largest and smallest
No. This does not work. Because a local max can be smaller than a local min.

8. Originally Posted by curvature
No. This does not work. Because a local max can be smaller than a local min.
But we are working with the extrema aren't we?

9. Originally Posted by janvdl
But we are working with the extrema aren't we?
a relative max==local max

10. Is it relative maximum: x=0; relative minima: x =+-1?

11. Originally Posted by Samantha
Is it relative maximum: x=0; relative minima: x =+-1?
$\displaystyle x = -1$ is the relative maxima.

$\displaystyle x = +1$ is the relative minima.

12. Originally Posted by janvdl
Besides, it's very clear to see here that we only have absolute maximum and minimum points.
Wrong! See the curve above. There is no absolute maximum nor absolute minimum points. Only relative or local ones.

13. Originally Posted by janvdl
$\displaystyle x = -1$ is the absolute maxima.

$\displaystyle x = +1$ is the absolute minima.

Im not sure what $\displaystyle x = 0$ is.
x=0 is a stationary point but not a extreme point because the sign of the derivative does not change there.

14. Originally Posted by curvature
Wrong! See the curve above. There is no absolute maximum nor absolute minimum points. Only relative or local ones.
Yes, i realised that now, my apologies Curvature.

Would you mind explaining the other method of calculating maxima and minima to me?

15. So there are no maximum nor minimum?

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