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Q&A

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

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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 $2<x<4$

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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 $x=2,4$.

enter image description here

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.

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