Thread: Critical Numbers

I need to find the critical numbers for the following problems:

1. f'(x) = 5x^4-6x^2+1
2. f'(x) = 2x^(-1/3)-1
3. f'(t) = -sin t, -2(pi)<=t<=2(pi)
4. f'(x) = (-2x(x+2))/(x-2)^3
5. f'(x) = e^x/(1+e^x)^2

I got an answer for the first few but I don't think it's corrects.

for 1:
0

for 2:
8

At this point I figured I was probably doing something wrong. Could someone please walk me through how to do these problems? I'd really appreciate it.

Hi, Critical numbers will be where your derivative equals 0.
You are lucky enough that you already have f'(x), which is the derivative of some function f(x).

So all you need to do is find what x's make f'(x) = 0

So for example number 1,
solve 0 = 5x^4 -6x^2 + 1

this is easily done by making the substitution u = x^2

so

0 = 5u^2 - 6u +1

Solve by using the quadratic equation:

u = (6 +/- sqrt(36 - 20)) /10

u = (6+/- 4)/10

so u = 1 or 1/5

since x^2 = u
x^2 = 1 or 1/5

if x^2 = 1 then x = +/-1

if x^2 = 1/5 then x = +/-sqrt(1/5)

So you have a total of four critical points for number 1,

x= {-1, -sqrt(1/5), sqrt(1/5), 1}

try the others like this and see if you can get em.

good luck.

I think I got the right answer for the first problem. But what do I do if there is only two variables.

Ex: The f''(x) of the first question is 20x^3-12x.

Do I factor out a 4x and set:
4x=0 and 5x^2-3=0?