Question

Help on finding the critical points of the following function with its first derivative

f'(x)=2e^x(x^2+25)(x^2-16)(x-3)^8
---------------------------
ln(x^2+100)

Answer 1

*

Answer 2

Taking the derivative of this thing is a pain. I'd use mathematica, personally.

Answer 3

http://www.wolframalpha.com/input/?i=derivative+of+%282e%5Ex%28x%5E2%2B25%29%28x%5E2-16%29%28x-3%29%5E8%29%2Fln%28x%5E2%2B100%29

Answer 4

This is already the first deriviative of a function

Answer 5

Im having trouble on ow to solve for X

Answer 6

Oh, I see.

Answer 7

Because that is how you find ciritcl points right?

Answer 8

The critical points are any x values that make the first derivative 0.

Answer 9

So if you set the derivative equal to 0, then solve for x, that should do it for ya.

Answer 10

It cant be a a whole number because the denominator is a log. It would have to be E or something along the lines of that

Answer 11

Oh, and it's also a critical point if the denominator is 0.

Answer 12

Not true. If the numerator is 0, then the whole thing will be 0.

Answer 13

0/anything = 0

Answer 14

Except for 0/0 which is undefined.

Answer 15

Ok. So I started off by multiplying by the denominator for both sides. Is that correct?

Answer 16

Sure. What equation did that yield you?

Answer 17

set \[2e^x(x^2+25)(x^2-16)(x-3)^8=0\]

Answer 18

as a fraction is only zero when the numerator is
then note that \(e^x\) is never zero
neither is \(x^2+25\) so it becomes much easier

Answer 19

\[\ln (x^2+100)= 2e^x(x^2+25)(x^2-16)(x-3)^8\]

Answer 20

???

Answer 21

Incorrect.

Answer 22

woops sorry. The left half hould be 0 because it is multiplied by 0

Answer 23

you are not trying to find where it is one, only where it is zero

Answer 24

Right.

Answer 25

forget the denominator, it is never zero, just concentrate on the numerator

Answer 26

Ok, I first divided by 2 on both sides

I got \[e^x(x^2+25)(x^2-16)(x-3)^8\]

Then i took the 8th root

I got \[e^x(x^2+25)(x^2-16)(x-3)\]

Answer 27

That actually doesn't work, algebraically speaking. You would have to take the 8th root of the whole side, not just the one term.

Instead, consider this fact. That side is just 4 things multiplied together. If any one of those 4 things is 0, then the whole side will be 0.

Answer 28

Now, recognizing that, just look at each of the 4 things and ask yourself when that thing will be 0.

Answer 29

Oh since (x^2-16 cant be 4, a critical point is 4

Answer 30

and 3

Answer 31

Well, I'm not sure what you mean by saying it "can't be 4" but if x=4, then (x^2-16) would be 0, so the whole thing would be 0.

Answer 32

Right. 3 is another critical point.

Answer 33

So 0/3/4 are all critical points. Wouldnt -4 also be one as well?

Answer 34

0 is actually not a critical point. I'm not sure why you just said it was.

Answer 35

But -4 is one.

Answer 36

Ok so how would I write this out if I am supposed t show ym steps? I just ignore the denominator and state that since if the numerator is 0, the whole answer will be 0?

Answer 37

Essentially. It's not hard to show that. 0/anything = 0

Answer 38

Or, if you want to do it algebraically, you can just do what you actually did on your own. Set the whole thing equal to 0, then multiply both sides by the denominator.

Answer 39

My professor is a real stickler when it comes to work. But since that I stated that 0/anything is always 0, he cant be a jerk about it.

Answer 40

If he's a stickler, just do what you did on your own. That's not anything he can complain about.

\(\Large \frac{2e^x(x^+25)(x^2-16)(x-3)^8}{ln(x^2+100)}=0\)
Multiply both sides by \(ln(x^2+100)\)

Answer 41

gives \(2e^x(x^2+25)(x^2-16)(x-3)^8 =0\)

Answer 42

Then divide by 2 which gives

\[e^x(x^2+25)(x^2-16)(x-3)^8\]

Answer 43

=0

Answer 44

Satellite73, could you help me solve this equation algebraically?