Question

A(t)=t^5-5t^4+20t^3-17 I am having trouble setting the derivative equal to zero. Please help!

Answer 1

What did you get for the derivative?

Answer 2

I've got
\[A' = 5t^4-20t^3 + 60t^2 = 5t^2(t^2-4t + 12)\]

Answer 3

THat's exactly what I've got

Answer 4

a*b=0 implies either a=0 or b=0 or both=0

Answer 5

Well I know that 5t^2=0 but I when I solve the parentheses for 0 (using quadratic formula) it comes out negative. I can't fit that into the rest of my problem!

Answer 6

(I am solving for the intervals of increasing and decreasing in the function)

Answer 7

you mean the number under the square root is negative?

Answer 8

Yes

Answer 9

so it is imaginary that means your only critical number is 0

Answer 10

That just means that that part of the equation doesn't have any real roots.

Answer 11

Okay. So if I am solving for the intervals, do I just divide my number in half?

Answer 12

I like using test points before and after each critical number

Answer 13

You're looking for intervals for what? Where it's increasing and decreasing?

Answer 14

Yes, @polpak

Answer 15

the derivative tells us the slope therefore tells us if is increasing or decreasing so choose a number before and after 0 to see what if it is increasing or decreasing

Answer 16

So it's increasing from negative infinity to infinity, correct?

Answer 17

The derivative is always positive except at 0. So it is increasing for \(x \in (-\infty,0) \bigcup (0,\infty)\)

Answer 18

It is not increasing at 0.

Answer 19

it has a horizontal tangent at x=0 thats what we found when we set A'=0

Answer 20

A'=0 means we are trying to find all x that has slope 0

Answer 21

Okay, thanks guys!