# Critical points of a function

Tags:
1. Oct 11, 2015

### Niaboc67

1. The problem statement, all variables and given/known data

1.) Let g(x) = x − (5/x^2)

find all critical numbers (if any) of g.

Give answers in increasing order (smallest first). Enter DNE in any unused space.

x =

x =

and

2.) Let g(x) = x |x + 5|

3. The attempt at a solution

I know I must first take the derivative and then set to zero, in order to find it's zero.

g(x) = x - 5/x^2
g'(x) = 1+10x^-3
g'(x) = 10/x^3 = -1
g'(x) = (x^3)10/x^-3 = -1(x^3)
g'(x) = 10 = -1(x^3)
g'(x) = -10 = x^3
g'(x) = -10^(1/3) = x
After here I don't know what I am suppose to do in order to find the crical points :/

Last edited by a moderator: Oct 12, 2015
2. Oct 11, 2015

### andrewkirk

Critical points are where the derivative is zero or undefined. You've found where it's zero. Now, for what value(s) of x is it undefined?

3. Oct 11, 2015

### RJLiberator

That seems to be right. You have a correct x value from my understanding. To find the point, you input x into the correct function to receive the y output.

edit: Andrew posted a better response.

4. Oct 12, 2015

### SammyS

Staff Emeritus
That is correct for the derivative, g'(x) .

After this you apparently meant to set g'(x) = 0 and then solve for x. However, you kept setting each line equal to g'(x).

None of the following lines is equal to g'(x).
There are many contexts for critical numbers.

Here it appears that you are trying to find places at which the derivative can change sign.

Essentially, you need to identify places where the derivative is zero and places where the derivative is undefined.

5. Oct 12, 2015

### Ray Vickson

Does that definition (non-existence of derivative) apply also at points where the function itself is undefined? That is the case here!

6. Oct 12, 2015

### Staff: Mentor

To elaborate on what SammyS said, the last correct line you (Niaboc67) wrote before the first line quoted above was
g'(x) = 1+10x^-3

The next step should be to set g'(x) = 0
$g'(x) = 0 \Rightarrow 1 + 10x^{-3} = 0$
Now, work with that.

7. Oct 12, 2015

### andrewkirk

I presume so, as the function being undefined at $x=a$ necessarily implies that the derivative is undefined there, even if $\lim_{x\to a}f'(x)$ exists.

To me it seems more natural to define a critical point as one where the derivative is zero or the function is undefined. A difference would be cases like $y=x^\frac{1}{3}$, which would be critical at at $x=0$ under the first definition but not under that alternative.

8. Oct 12, 2015

### ehild

The critical point must belong to the domain of the function. Think of the function g(x)=√x. Are all x<0 critical points? x=0 is the only critical point, as the function exists there but its derivative does not exist.

We say that [PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/eq0001M.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/empty.gif [Broken] is a critical point of the function f(x) if f(x) exists and if either of the following are true: f '(c) =0 or f '(c) does not exist.

[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/empty.gif[PLAIN]http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints_files/empty.gif [Broken]

Last edited by a moderator: May 7, 2017