Critical Number of a Function

Question

Please find all of the critical numbers of the function f(x)=2x^3+3x^2-12x

step by step locate all relative extrema, and the open intervals on which the function is increasing or decreasing.

Answer 1

To find the critical values, find the first derivative and set it equal to zero. Solve for those values.

f'(x)=6x²+6x-12 Factor out the GCF

f'(x)=6(x²+x-2)

f'(x)=6(x+2)(x-1)

x+2=0

x=-2 critical value

x-1=0

x=1 critical value

Next, need to set up a table to use test points to determine if the slope is positive (increasing) or negative (decreasing) At the critical value if the slope changes from increasing to decreasing, it is a relative max. If the slope changes from decreasing to increasing, it is a relative min.

x=-3 6(-3+2)(-3-1) = positive slope, increasing

x=-2 CV relative max

x=0 6(0+2)(0-1) = negative slope, decreasing

x=1 CV relative min

x=2 6(2+2)(2-1) = positive slope, increasing

Using the x value, plug them in to find the y values for the min and max

relative max (-2,20)

relative min (1,-7)

increasing (-∞,-2), (1,∞)

decreasing (-2,1)

Answer 2

The easiest way to answer this question is to make a graph...a rough sketch will do or use your graphing calculator.

The critical points occur where the derivative = 0.

6x² + 6x - 12 = 0...x=-2 and 1

x=0 is a root.

With your sketch and this information you should be able to answer the rest of the question.