max/min

**Samantha** · July 26th 2007, 05:50 AM

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

What is relative maximum and relative minimum of it?

**janvdl** · July 26th 2007, 05:57 AM

Originally Posted by Samantha

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

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.

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

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

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

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

**curvature** · July 26th 2007, 05:58 AM

Originally Posted by Samantha

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

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.

**Samantha** · July 26th 2007, 06:06 AM

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.

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

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

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

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

I don't understand =(

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

**janvdl** · July 26th 2007, 06:08 AM

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

**curvature** · July 26th 2007, 06:17 AM

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.

**curvature** · July 26th 2007, 06:18 AM

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.

**janvdl** · July 26th 2007, 06:20 AM

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?

**curvature** · July 26th 2007, 06:22 AM

Originally Posted by janvdl

But we are working with the extrema aren't we?

a relative max==local max

**Samantha** · July 26th 2007, 06:41 AM

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

**janvdl** · July 26th 2007, 06:47 AM

Originally Posted by Samantha

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

$x = -1$ is the relative maxima.

$x = +1$ is the relative minima.

**curvature** · July 26th 2007, 06:47 AM

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.

**curvature** · July 26th 2007, 06:51 AM

Originally Posted by janvdl

$x = -1$ is the absolute maxima.

$x = +1$ is the absolute minima.

Im not sure what $x = 0$ is.

x=0 is a stationary point but not a extreme point because the sign of the derivative does not change there.

**janvdl** · July 26th 2007, 06:52 AM

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?

**Samantha** · July 26th 2007, 06:53 AM

So there are no maximum nor minimum?

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