Find the intervals where f is decreasing and concave up for f(x)=x^3-3x^2-24x+3. Answer with (a,b) or (a,b) U (c,d) for a < c.

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You need to remember that the first integral of the function tells if the function increases or decreases over an interval.

`f'(x) = 3x^2 - 6x - 24`

You need to find the zeroes of derivative such that:

`3x^2 - 6x - 24 = 0 =gt x^2 - 2x - 8 = 0`

You need to apply quadratic formula:

`x_(1,2) = (2+-sqrt(4 + 32))/2 =gt x_(1,2) = (2+-sqrt36)/2`

`x_(1,2) = (2+-6)/2 =gt x_1 = 4; x_2 = -2`

The derivative has negative values between (-2;1). The derivative has positive values over intervals `(-oo,-2) U (1 ; +oo).`

The function decreases if the derivative is negative, hence the function decreases over interval (-2;1).

The function increases if the derivative is negative, hence the function decreases over interval `(-oo,-2) U (1 ; +oo).`

You need to differentiate the first derivative with respect to x such that:

f"(x) = 6x - 6

You need to find the root of f"(x) such that: 6x - 6 = 0 => x = 1

You should notice that the second derivative is negative over the interval `(-oo, 1)`  and it is positive over `(1: +oo).`

The negative values of the second derivative tell you that the graph of function is concave down.

Hence, the function decreases over interval (-2;1) and its graph is concave up over interval `(1;+oo).`
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