Finding the critical numbers

**endiv** · November 15th 2008, 03:10 PM

Find critical numbers for x^(4/5) * (x-1)^2?

If you could help out, please. I've already done most of the work, but I'm not too sure on where to go and how to work with the problem to get the critical numbers.

(1). The derivative of the problem. Using the product rule you get

x^(4/5) * 2(x-1) + (x-1)^2 * 4/5x^-1/5

here's where I'm stuck.. the factoring

Relating to the solutions manual for a similar problem ( x^(4/5) * (x-4)^2 ), I know that I'll be able to factor it by:

1/5x^-1/5 * (x-1)

This is where I'm stuck... factoring will I get:

(1/5)x^-(1/5)*(x-1) [ 5 * x * 1 + (x-1) *4] ????

can anyone give me a push on this? I should be able to solve this step after the factoring.

-Thanks

**Soroban** · November 15th 2008, 03:26 PM

Hello, endiv!

There's some tricky algebra in this one . . .

Find critical numbers for: . $f(x) \:=\:x^{\frac{4}{5}}(x-1)^2$

Differentiate and equal to zero . . .

. . $f\,'(x) \;=\;x^{\frac{4}{5}}\cdot2(x-1) + \tfrac{4}{5}x^{-\frac{1}{5}}(x-1)^2 \;=\;0$

We have: . $2x^{\frac{4}{5}}(x-1) + \frac{4(x-1)^2}{5x^{\frac{1}{5}}} \;=\;0$

Multiply by $5x^{\frac{1}{5}}\!:\;\;10x(x-1) + 4(x-1)^2 \;=\;0$

Factor: . $2(x-1)\bigg[5x+2(x-1)\bigg] \;=\;0 \quad\Rightarrow\quad 2(x-1)(7x-2) \;=\;0$

Therefore, the critical values are at: . $x \;=\;1,\:\tfrac{2}{7}$

**endiv** · November 15th 2008, 03:57 PM

Originally Posted by Soroban

Hello, endiv!

There's some tricky algebra in this one . . .

Differentiate and equal to zero . . .

. . $f\,'(x) \;=\;x^{\frac{4}{5}}\cdot2(x-1) + \tfrac{4}{5}x^{-\frac{1}{5}}(x-1)^2 \;=\;0$

We have: . $2x^{\frac{4}{5}}(x-1) + \frac{4(x-1)^2}{5x^{\frac{1}{5}}} \;=\;0$

Multiply by $5x^{\frac{1}{5}}\!:\;\;10x(x-1) + 4(x-1)^2 \;=\;0$

Factor: . $2(x-1)\bigg[5x+2(x-1)\bigg] \;=\;0 \quad\Rightarrow\quad 2(x-1)(7x-2) \;=\;0$

Therefore, the critical values are at: . $x \;=\;1,\:\tfrac{2}{7}$

holy.. wow... I didn't think of it that way! You even did the extra of giving the critical numbers. (embarassed)

Thanks so much!

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