Question

Determine the open intervals on which the given function is increasing or decreasing and the coordinates of any relative extrema.

Answer 1

\[f(x)=x ^{2/3}-4\]

Answer 2

I took the derivative and set it equal to zero, but when I solved it I got 0 not equal to 2 not sure what I am doing incorrect

Answer 3

\[f' (x)=\frac{ 2 }{ 3\sqrt[3]{x} }\]

Answer 4

0 not equal to 2?
I'm not sure what that means :o

Answer 5

\[\frac{ 2 }{ 3\sqrt[3]{x} }=0\]

Answer 6

Your derivative looks good.
Looking for critical points,\[\Large\rm 0=\frac{2}{3}x^{-1/3}\]Multiply each side by 3/2,\[\Large\rm 0=x^{-1/3}\]So it looks like we get a critical point at x=0, yes? :o

Answer 7

boy do I feel dumb today

Answer 8

now I know how to finish this problem
can you help me solve one more?

Answer 9

sure c:

Answer 10

\[h(x)=\frac{ x^2-3x-4 }{ x-2 }\]

Answer 11

same directions
so I took the derivative and got

Answer 12

\[h'(x)=\frac{ x^2-4x+10 }{ (x-2)^2 }\]

Answer 13

so is my critical point x=2 or I thought I set it equal to zero and solve for x? any case that meant I have to solve the numerator and I got a complex solution so what am I overlooking again?

Answer 14

We deal with the numerator and denominator separately.
The denominator is giving us critical points for a different reason than we're used to.
It's not giving us locations of horizontal slope.
This was true for the last problem as well, the critical point ended up being a cusp.

The problem works out a little differently, you find x=2 to be a critical point.
But wait! We have to ask ourselves, "Is x=2 in the DOMAIN of my function?"

Answer 15

so the last problem does not have a relative extrema because x=0 is not differentiable.
and you are right x =2 is not in the domain

Answer 16

does not have relative extrema because the function is not differentiable at x=0,
Ooo yah I like the way you said that! :O

The last one `did` have a critical point, but not the kind we were hoping for >.<

Answer 17

so the critical point is complex? or in this case undefined

Answer 18

but what I am to write, I am lost on this one

Answer 19

For the second one?
No critical points exist within the domain of the function, therefore no extreme points exist either.

Or however you think that should be worded.
Oh oh oh, but you still need to determine increasing/decreasing, yes?

Answer 20

yes, should I test point on both sides of 2 since I know that x=2 is the vertical asymptote and is an undefined point?

Answer 21

Yah that seems like a good idea >.<

Answer 22

thanks, I couldn't think of anything else.....thank you so much

Answer 23

only 500 more impossible calculus problems to go.....

Answer 24

Try to simplify things if you can.
Since we only care about the `sign` of the result,
notice that the denominator is always positive, so it's not telling you anything:\[\Large\rm h'(x)=\frac{ x^2-4x+10 }{ positive }\]And then plug the `less than 2` or `greater than 2` value into the top to see what's going on.

Answer 25

Impossible?? You mean super funnnn? \c:/

Answer 26

yes, I try to do them as best I can and I come here to double check my solutions, thoughts, process, etc........