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

[Allasolid]

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.

 

[Astronuc]

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.}

 

[Allasolid]

Oh here's the derivative i need to solve -e^x(3x⁵+x⁴-34x³+55x²+169x-216)/(x²-9)²

 

[Ray Vickson]

Oh here's the derivative i need to solve -e^x(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

 

[Allasolid]

Good point. Sorry I am new to this.

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

 

[SammyS]

Good point. Sorry I'm new to this.

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

Oh here's the derivative i need to solve -e^x(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.

 

[Allasolid]

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.

 

[HallsofIvy]

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 $3x^5+ x^4- 34x^3+ 55x^2+ 169x- 216= 0$.