Question

Find the local extrema for f'(x)= x(x-3)^2 and classify as local maxima or minima

Answer 1

To find the maxima or the minima you would need to find the derivative of the function and set it equal to 0.
That is so you can find when the slope of the original function = 0, giving you the x value for when there might be a maxima or minima
To find the derivative, you'd need to do the product rule & the power rule
After finding the x coordinates, make a sign chart, and plug in points to the derivative on both sides of the point, giving you whether the slope is positive or negative, thus determining whether it's a minima or maxima

Answer 2

An alternative to making a sign chart is just taking the derivative of f'(x).

This will give you the second derivative and will tell you the curvature of the original function. If the f''(x) is positive then it's a minimum and if f''(x) is negative then the point is a maximum.

Answer 3

@CanadianAsian so i find the derivative first, so for this equation for example:
y= 3x^2+24x-8
y'=6x+24
y'= 6(x+4)
0=6(x+4)

What do i do after this ?

Answer 4

So now that you know that at x=-4 the slope of the original function is 0, you know it's either a minimum or maximum right?
You can make a sign chart

Answer 5

This tells us that on the left of -4, the function has a negative slope (you can find this by plugging in a value on the left of -4 into the derivative, like -5)
And that there is a positive slope on the right of -4.
Since the function goes from negative to positive, it has a minima

Answer 6

The way CanadianAsian does it is taking another derivative giving you the concavity of the original function which you would find to be 6
Since 6 is positive, the original function is concave up, which means you have a minima

Answer 7

I think i was supposed to do it like both ways, to find if it concaves, up or down and also the local max/min. Thank you :)