minima/maxima: f(x) = x^4(x - 3)^5 for x in [-11, 13]

[littlejodo]

For x (domain? funny e-like symbol) [-11,13] the function f is defined by f(x) = x^4(x-3)^5.

I am supposed to find where the function is increasing and where the function is positive and where it is at its minimum, which I think I could do if I could just get the critical points.

f'(x) = 4x^3(x-3)^5 + x^4 * 5(x-3)^4(x) --- product rule and chain rule

f'(x) = 20x^8(x-3)^9 = 0 --- combine terms

So now I need to solve for x. This is where I am stuck. I don't want to have to multiply out all of the (x-3)s because I figure that there has to be a simpler way... only, I cannot figure out what that is. Simple algebra probably, but I need some help.

If anyone can point me in the right direction here, I would really appreciate it!

 

[daon]

Re: Lost in the algebra of minima/maxima

f'(x) = 4x^3(x-3)^5 + x^4 * 5(x-3)^4(x) --- product rule and chain rule

f'(x) = 20x^8(x-3)^9 = 0 --- combine terms

You didn't do algebra correctly here...

 

[littlejodo]

Re: Lost in the algebra of minima/maxima

what about this:

x^4(x-3)^5

f ' (x) = 4x^3(x-3)^5 + 5x^4(x-3)^4(x) = 4x^3(x-3)^5 + 5x^5(x-3)^4

Is that correct? If so, is there an easier way to solve for x than to multiply everything out?

 

[daon]

Re: Lost in the algebra of minima/maxima

is there an easier way to solve for x than to multiply everything out?

Yes, factor. Your chain rule application was also wrong on the second term. If u = x-3, u' = 1.