Max and Min Values

1. Nov 16, 2005

Quadruple Bypass

OK, the professor did this problem for us as an example and i got lost somewhere in between. the problem was: find the critical points
f(x)= x^(4/5)*(x-4)^2

then he used the product rule to get

f '(x)= 4/5 x^(-1/5) *(x-4)^2 + x^(4/5) * 2(x-4) = 0

THEN, the part that threw me off was the next part where he said multiply both sides by 1/5....what did he mean by that? after that you get

4/5 (x-4)^2 + 2x(x-4) = 0 and so on...

my question is how and what did he do to get rid of the x^(-1/5) and the x^(4/5)???

Any help would be greatly appreciated

2. Nov 16, 2005

NateTG

Multiply both sides by
$$x^{\frac{1}{5}}$$

3. Nov 16, 2005

shmoe

He multiplied both sides by $$x^{1/5}$$, not 1/5.

4. Nov 16, 2005

Hammie

Don't multiply by 1/5, try multiplying by x^(1/5). The whole idea is to turn the exponents into integers, and then you can use the usual methods to find the roots of the equation.

Last edited: Nov 16, 2005
5. Nov 16, 2005

NonSequitur

He is not multiplying by 1/5, he is multiplying by x^(1/5). On the zero side, it of course goes away. On the other side, it distributes and x^(1/5)*x^(-1/5)=1 while x^(1/5)*x^(4/5)=x nicely killing off those pesky rational exponents. Good Luck

6. Nov 16, 2005

Quadruple Bypass

WOOHOOO!!!!! Thanks very much!

7. Nov 16, 2005

HallsofIvy

Staff Emeritus
"Never seen such unanimity of opinion before in my life"
Poobah, in "The Mikado"

8. Nov 17, 2005

Hammie

And who said mathmaticians don't have a sense of humor??

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