# 2nd derivative test of a polynomial and related material

1. Mar 28, 2009

### montana111

1. The problem statement, all variables and given/known data
im working on developing the graph for the following function: h(x)= (x+1)^5 -5x -2.
i have h'(x) = 5(x+1)^4 -5
and h''(x) = 20(x+1)^3

i get stuck when trying set h'(x) = 0. i know that there are four roots because i graphed it on my calculator but using traditional methods i can only figure out mathematically 2 of them i.e. 1 and -1. my calculator shows the other two at approx .45 and -.45. can someone please guide me to the simplest way to break down h'(x) and find its roots? thank you.

2. Relevant equations

h'(x) = 5(x+1)^4 -5

3. The attempt at a solution

ive tried many different methods to reach a solution,1) factoring (couldnt figure it out), 2) trying to get each side down to x^2 factors so i could use the quadratic formula (got messy/didnt work correctly *i think!*) 3) just looking at it and plugging in numbers (this is how i got -1 and 1). I just dont know what to do aaaaaannnd the answer is not in the back of the book.

2. Mar 28, 2009

### Asteroid1

Let's start with saying that you were able to take the derivatives correctly.

If I understood it correctly you wanted to find the x for where the first derivative equals 0. Here is where you started typing down things that didn't make much sense. You say you filled in the -1 and the +1 for the x. Those numbers aren't giving me a 0 at the end.
$$5(x+1)^{4}-5=5(-1+1)^{4}-5=5(0)^{4}-5=0-5=-5$$ For the x=-1

Are you sure you were working with the right formula and numbers?

As a tip: Try solving for $$(x+1)^{4}$$

3. Mar 29, 2009

### Staff: Mentor

The equation h'(x) = 0 has only one root, of multiplicity 4. Apparently you entered the equation incorrectly in your calculator, or else it's no good and you should throw it away.

Solve the equation 5(x + 1)4 - 5 = 0 -- it's an easy one to solve. I can do it in my head without a graphing calculator.

4. Mar 29, 2009

### the_awesome

h'(x) = 5(x+1)4 -5

when x = -2, h'(x) = 0
when x = 0, h'(x) = 0

5. Mar 29, 2009

### HallsofIvy

Staff Emeritus
??? No, it has 4 distinct roots, two real, two imaginary. Since you are graphing this over the real numbers, you only wnat the two real roots.

Let u= x+ 1 and first solve u4= 1. Again, that has 4 (easy) roots, two real, two imaginary. Then solve x+ 1= u for x.