F(x)=2x^3+3x^2-36x+20

Question

Find all the intervals for which f(x) is increasing, decreasing and concavity. I need the answers but I need all the steps too.

Answer 1

Set F'(x) equal to zero.  This is the first derivative test.

 

6x² + 6x - 36 = 0

 

6(x² + x - 6) = 0

 

x² + x - 6 = 0

 

(x + 3)(x - 2) = 0

 

x = -3        and          x = 2

 

These are your critical values.  Since this is a cubic function with a positive coefficient, we can picture the general shape of this graph.  Knowing this fact, we don't have to evaluate the derivative using test points around these critical values.

 

Intervals of increase: (-∞, -3)∪(2, ∞)

Intervals of decrease: (-3, 2)

 

 

Now we use 2nd derivative test to find concavity.  Second derivative equal to zero.

 

12x + 6 = 0

 

6(2x + 1) = 0

 

x = -1/2

 

This is where the point of inflection occurs.

 

If we evaluate the derivative before and after x=-1/2, we can see where the 2nd derivative is positive and negative.  If positive, then concave up.  Otherwise, concave down.

 

F"(-1) = 6(-1) = -6

F"(0) = 6(6) = 6

 

Concave down in the interval (-∞, -1/2)

Concave up in the interval (-1/2, ∞)