How many critical points does the function have?
f(x)=(x-2)^2(x+3)^2

Question

How many critical points does the function have?
f(x)=(x-2)^2(x+3)^2

arbershabani97 · Answer

two,  those are 2 and -3

anonymous · Answer

@arbershabani97 

One would need to find f ' (x) and set equal to zero to determine the critical points. 

2 and -3 are zeros of the function, but not necessarily critical points of f(x).

ganeshie8 · Answer

they're also critical points, as the multiplicity is 2, 
however you still need to find derivative and see if there are other critical points

anonymous · Answer

It could very well be they are critical points, but the student may assume that the zeros are the critical points, since no work was shown, or no mention of the derivative was mentioned.

ganeshie8 · Answer

agree dude, to figure out that they're critical points you dont need derivative work.

to see if other critical points exist or not, you need to solve f'=0. i was saying that ok

anonymous · Answer

@YoselinReyes 

To find crtitical points of a function, take the derivative of f(x), f ' (x), and set equal to zero or where the derivative is undefined. Those will be the critical points of the function f(x).

anonymous · Answer

yea because the answer it says that there are three critical point and did figure that -3 and 2 were the first two so if I did find the derivative would the answer be what I get from setting it to zero or would zero be the third critical point

anonymous · Answer

Yes, just set derivative equal to zero or find where the derivative is undefined (which doesnt apply in this case). Don't ever look at a function and determine its critical points "by looking".

anonymous · Answer

Functions can be very deceiving by "its looks".

anonymous · Answer

well yea i know that but if I did find the derivative would the answer be what I get from setting it to zero or would zero be the third critical point

anonymous · Answer

If when setting the derivative equal to zero, you get x = 0 as a solution, then x = 0 is also a critical point.

Any solution to f ' (x) = 0 is a critical point!

anonymous · Answer

Ohhhh ok thanks

anonymous · Answer

Welcome. And I wanted to make that point crystal clear, never ever look at a function and think that its zeros are automatically critical points. That is an absolute false statement.

anonymous · Answer

They may or may not be. Finding the derivative is the way it is defined in calculus.