2nd derivative test of a polynomial and related material

[montana111]

Homework Statement

I am 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.

Homework Equations

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

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 don't know what to do aaaaaannnd the answer is not in the back of the book.

 

[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}

 

[Mark44]

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.

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.

 

[the_awesome]

h'(x) = 5(x+1)⁴ -5

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

 

[HallsofIvy]

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.

? 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.