Stuck on exponents

[Mayuko]

The problem says: find the first derivative and use it to find the increasing and decreasing intervals and any relative extrema.

f(x)= x^2/3 -2x^1/3

I found the first derivative: f '(x)= (2/3x^-1/3) - (2/3x^-2/3)

Then you set it equall to zero: (2/3x^-1/3) - (2/3x^-2/3) = 0 ; then solve for x.

That's where I'm stuck. I cant figure out how to find x. Where should I start?

Thanks for the help.

 

[Deleted member 4993]

The problem says: find the first derivative and use it to find the increasing and decreasing intervals and any relative extrema.

f(x)= x^2/3 -2x^1/3

I found the first derivative: f '(x)= (2/3x^-1/3) - (2/3x^-2/3)

Then you set it equall to zero: (2/3x^-1/3) - (2/3x^-2/3) = 0 ; then solve for x.

That's where I'm stuck. I cant figure out how to find x. Where should I start?

\(\displaystyle \frac{2}{3}x^{\frac{-1}{3}} - \frac{2}{3}x^{\frac{-2}{3}} \, = \, 0\)

\(\displaystyle \frac{2}{3}x^{\frac{-1}{3}} (1\, - \, x^{\frac{-1}{3}} ) \, = \, 0\)

Now continue.....

Thanks for the help.

 

[Mayuko]

The problem says: find the first derivative and use it to find the increasing and decreasing intervals and any relative extrema.

f(x)= x^2/3 -2x^1/3

I found the first derivative: f '(x)= (2/3x^-1/3) - (2/3x^-2/3)

Then you set it equall to zero: (2/3x^-1/3) - (2/3x^-2/3) = 0 ; then solve for x.

That's where I'm stuck. I cant figure out how to find x. Where should I start?

\(\displaystyle \frac{2}{3}x^{\frac{-1}{3}} - \frac{2}{3}x^{\frac{-2}{3}} \, = \, 0\)

\(\displaystyle \frac{2}{3}x^{\frac{-1}{3}} (1\, - \, x^{\frac{-1}{3}} ) \, = \, 0\)

Now continue.....

Thanks for the help.

2/3x^-1/3=0
3/2 (2/3x^-1/3)=0 (3/2)
x^-1/3=0
1/(x^1/3)=0
(x^1/3)/1 * 1/(x^1/3)=0 *(x^1/3)/1
1 = 0

(1-x^-1/3) = 0
-x^-1/3 = -1
1/(x^1/3) = 1
(x^1/3)/1 * 1/(x^1/3)= 1 *(x^1/3)/1
1 = x^1/3
1 = cube root of x
1^3 = cube root of x cubed
1 = x

That's what I got. Does it seem right?

Thanks again for your help.

 

[Deleted member 4993]

The problem says: find the first derivative and use it to find the increasing and decreasing intervals and any relative extrema.

f(x)= x^2/3 -2x^1/3

I found the first derivative: f '(x)= (2/3x^-1/3) - (2/3x^-2/3)

2/3x^-1/3=0
3/2 (2/3x^-1/3)=0 (3/2)

Simpler steps...

\(\displaystyle \frac{2}{3}x^{-\frac{1}{3}} = 0\)

\(\displaystyle x^{-\frac{1}{3}} = 0\)

\(\displaystyle \text{no solution}\)


(1-x^-1/3) = 0
-x^-1/3 = -1
1/(x^1/3) = 1
(x^1/3)/1 * 1/(x^1/3)= 1 *(x^1/3)/1
1 = x^1/3
1 = cube root of x
1^3 = cube root of x cubed
1 = x <<<< Correct

That's what I got. Does it seem right?

Thanks again for your help.

 

[galactus]

\(\displaystyle \frac{d}{dx}[x^{\frac{2}{3}}=\frac{2}{3}x^{\frac{-1}{3}}\)

\(\displaystyle \frac{d}{dx}[2x^{\frac{1}{3}}]=\frac{2}{3}x^{\frac{-2}{3}}\)

So, we have \(\displaystyle \frac{2}{3}x^{\frac{-1}{3}}-\frac{2}{3}x^{\frac{-2}{3}}=\frac{2(x^{\frac{1}{3}}-1)}{3x^{\frac{2}{3}}}\)

To find the extrema, set to 0 and solve for x. After factoring, the solution is obvious?.

If \(\displaystyle f'(x)>0\) for every value of x in (a,b), then f is increasing in the interval [a,b].

If \(\displaystyle f'(x)<0\)for every value of x in (a,b), then f is decreasing in the interval [a,b].

If \(\displaystyle f'(x)=0\)for every value of x in (a,b), then f is constant on [a,b].

Graph it. That will help you see how the slope runs. Check the interval between 0 and 1 and then from 1 on up.

Try, say, x=1/2 in f'(x). Is it >0 or <0?. Well, if we use x=1/2, we get \(\displaystyle f'(1/2)= -0.21832\)

Therefore, it is decreasing.