Find the intervals on which it is increasing and those on which it is decreasing of the following function.

Question

Find the intervals on which it is increasing and those on which it is decreasing of the following function. $$f(x)=x^3-9x^2+24x-12,0\leq x\leq 6$$

After differentiating (once) the function I get that $x=2,4$. But, I was getting confusing by reading those description of in some intervals it is increasing in some intervals it is decreasing.

Here, symbol of $f'(x)$ is $=(-)(-)=+$ (positive) in $0\leq x<2$ interval. that means $f'(x)>0$. Hence, the function is increasing in the interval $0\leq x<2$.

I was confused for symbol/sign. They wrote $(-)(-)=+$ (positive) but, where they found those sign? If I can understand for the interval than, I can understand for other intervals also. That's why I didn't add further information.

I guess most of people may confuse for a single description. So, I am adding further information from book again.

symbol $f'(x)$ is $=(+)(-)= negative$. That means $f'(x)<0$, hence the function is decreasing in the interval $2<x<4$

Answer 1

I was watching the tutorial. When I differentiated that function (which I wrote in question). I got

$$f'(x)=3x^2-18x+24$$

When I worked on the equation further. And, put that $f'(x)=0$ than, I got that $x=2,4$.

I wrote down intervals then.

$$0\leq x<2$$ $$2<x<4$$ $$4<x\leq6$$

I was putting values (from interval) in function then.

$$f'(x)=3(0)^2-18(0)+24$$ $$=24$$ So, the value is increasing the interval ( $[0,2)$ ) since it is positive value.

$$f'(x)=3(3)^2-18(3)+24$$ $$=-3$$

So, the value is decreasing. I will get positive value for $5$ and $6$. For that reason that is increasing.

Answer 2

The derivative can be written as $f'(x)=3(x-2)(x-4)$. If you plug in a value $x<2$, then both parenthesis will be negative, so the product will be positive. That is probably what they mean with $(-)(-)=+$, but note that this is very informal notation.