Thread: Finding Intervals of derivatives including local max or min

I have a function f(x) = x^3 - 3x^2 + 1. Using the First derivative:

1. Identify the intervals over which f(x) is increasing and decreasing. Hence classify the turning points (as local maximum or local minimum).

using the first derivative i can get to: f ' (x) = 3x^2 - 6
f ' (x) = 0, f ' (x) = 3(x^2 - 2x)
->3(x-1)(x-1)-1 = 0
how do i proceed from here to get local max and local min? do i take the critical points as x=0, x=1, x=1?


2. Find any inflection points, and confirm your conclusion using what you know about the second derivative and it's relationship to concavity.

for this question i got the first derivative f ' (x) = 3x^2 - 6 then the 2nd derivative f '' (x)=6x - 6
where 6x-6 = 0 will give me x=1

now taking the no. larger than 1 and smaller than 1 and plugging it back in the 2nd derivative
-by plugging no. > 1 in the 2nd derivative ill get a positive no. hence Concave up
-by plugging no < 1 in the 2nd derivative ill get a negative no. hence Concave down

so the interval of Concave up is (-ve infinity , 1) and the interval for Concave down is (1, +ve Infinity)
is this correct?

You have differentiated the function incorrectly...you should expect at most two distinct critical values from a cubic function.

Although now I see the first expression you give for the derivative is probably a typo, as you do state it correctly as:

$\displaystyle f'(x)=3(x^2-2x)$

After this you restate it incorrectly.

Factor it further, to find your 2 critical values...