calculus question?

Question

Ok, could somebody tell me if I did this question right?

The question is:
The derivative of f(x) is f'(x)=6x(2x-5)^2(3x+8)^3. Using interval notation, tell where f(x) is increasing and where it is decreasing.

So I solved 6x(2x-5)^2(3x+8)^3=0 and got the critical values of x=0, x=5/2, and x= -8/3. Are these right? I know this should be simple algebra but im blankinggg.

Next I found that f(x) is decreasing on the interval (-infinity,o] and increasing on [0, +infinity).

Is this right? Thanks to anyone who can help.

Answer 1

f(x) is increasing as long as f '(x) is positive. At max or mins
f(x) is stationary. If f '(x) is negative, the f(x) is decraesing.
You can see that 6x and (3x+8)^3 and (2x-5)^2 are always positive if x>0. So f(x) is increasing for x>0.

In the interval (-8/3,0) f'(x) is negative because the 6x term makes it negative an hence f(x) is decreasing in this interval.

Finally f'(x) is positive for all x in the interval (-infinity, -8/3)
so f(x) is increasing in this interval.

Answer 2

That is a funny looking derivative to find critical values for.

The way I read your derivative you have a base of

6x(2x-5) which is being raised to this power: 2(3x+8)^3
[and this exponent itself has a base raised to the 3rd power].

Is that what you intended for your derivative??