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?
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Now you have to address inequality That is equivalent to factoring you conclude it by separating cases
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 . You'll find that this takes a zero at an . Between these zero , otherwise it's negative.
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
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 ?
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
As the product is greater than zero we have two possibilities 1) We conclude that 2) We conclude that Then the interval is
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