1. interval for f'(x)>0

the question is
on what interval is f'(x)>0 for f(x)=2x^3 - 3x^2 - 36x +2

i found f'(x) = 6x^2 - 6x -36

then what do i have to do next to find interval?

2. Now you have to address inequality $6x^2-6x-36 > 0$

That is equivalent to $x^2-x-6 > 0$

factoring $(x-3)(x+2) > 0$

you conclude it by separating cases

3. Originally Posted by haebinpark
the question is
on what interval is f'(x)>0 for f(x)=2x^3 - 3x^2 - 36x +2

i found f'(x) = 6x^2 - 6x -36

then what do i have to do next to find interval?
Factor $6(x-3)(x+2)$. You'll find that this takes a zero at $x = -2$ an $x = 3$. Between these zero $6(x-3)(x+2) >0$, otherwise it's negative.

4. f'(x) = 6x^2 - 6x -36

You should notice f ' is a parabola opening up. Locate the zeroes then f ' is negative between the zeroes and positive otherwise

Set f ' = 0

6x^2 - 6x -36 = 0

x^2 -x -6 = 0

Now factor to find the zeroes

5. so

f'(x) = 0 when x = 3 and x= -2

f'(x) > 0 when x > 3 and x < -2

am i able to write answer like -2<x<3 or should i leave it as f'(x) > 0 when x > 3 and x < -2 ?

6. Actually you would write x > 3 OR x < -2

as you are taking the union of (-inf, -2) U (3,inf)

-2 < x < 3 is where f ' < 0

at least you didn't write 3 < x < -2 like a lot of students would which is
nonsense

7. As the product is greater than zero we have two possibilities

1) $x-3 >0 \wedge x+2 > 0$

We conclude that $x \in (3,\infty)$

2) $x-3 <0 \wedge x+2 < 0$

We conclude that $x \in (-\infty, -2)$

Then the interval is $x \in (-\infty,-2)\cup (3,\infty)$