What are the critical points of a given function and how can they be solved?

Allasolid
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Homework Statement


I need to find the critical points of this functionhttp://www4a.wolframalpha.com/Calculate/MSP/MSP84719i6a1iabg11598d000049hhhie62iif1c81?MSPStoreType=image/gif&s=16&w=140&h=46

The Attempt at a Solution


I'm having trouble solving the first derivative, which is http://www4a.wolframalpha.com/Calculate/MSP/MSP12919i6a079bg6ed96c00003ed60iee4d866bb3?MSPStoreType=image/gif&s=39&w=493&h=48 . Well more specifically solving the 5th degree polynomial. It needs to be solved without using a graphing calculator.
 
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Please write out the function/polynomial.

Use Latex or text with [ /sup] tags as appropriate. Remove the space between [ /sup when using the superscript tag.
 
Oh here's the derivative i need to solve -ex(3x5+x4-34x3+55x2+169x-216)/(x2-9)2
 
Allasolid said:
Oh here's the derivative i need to solve -ex(3x5+x4-34x3+55x2+169x-216)/(x2-9)2

You should show the function itself as well. How do we know your derivative is correct?

RGV
 
Good point. Sorry I am new to this.

The function is (3x3+4x2-7x+17)/(ex(x2-9))
 
Allasolid said:
Good point. Sorry I'm new to this.

The function is (3x3+4x2-7x+17)/(ex(x2-9))

Allasolid said:
Oh here's the derivative i need to solve -ex(3x5+x4-34x3+55x2+169x-216)/(x2-9)2
That derivative is correct.

Unfortunately, the polynomial in the numerator does not appear to factor easily. The denominator is factorable.
 
Ohhhhh so i now have (x+3)(x-3)(x+3)(x-3) in the denominator.. :/ sorry i don't see how that's suppose to help.
 
Unless it cancels things in the numerator (and you can check that by seeing if x= 3 or x= 3 will make the numerator equal to 0), factoring the denominator is irrelevant. A fraction, a/b, is equal to 0 only if a= 0. You don't need to worry about the denominator. It is also true that e^x is never 0 so it all reduces to solving 3x^5+ x^4- 34x^3+ 55x^2+ 169x- 216= 0.
 
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