How do you determine whether the function #F(x)= 1/12X^4 + 1/6X^3-3X^2-2X+1# is concave up or concave down and its intervals?

Question

Emily Ammerman · Accepted Answer

To find intervals on which the graph of #F# is concave up and those on which it is concave down, investigate the sign of the second derivative.

For, #F(x) = 1/12x^4+1/6x^3-3x^2-2x+1# we have #F'(x) = 1/3x^3+1/2x^2-6x-2# and #F''(x) = x^2+x-6 = (x+3)(x-2)# The only chance #F''(x)# has to (perhaps) change sign is at #F''(x) = 0#. Which happens at #x=-3# and at #x=2# On #(-oo,-3)#, both factor of #F''# are negative, so #F''# is positive and the graph of #f# is concave up. On #(-3,2)#, we get #F''(x)# is negative, so the graph is concave down. On #(2,oo)#, #F''(x)# is positive, so the graph is concave up. The graph of #F# is concave up on #(-oo,-3)# and on #(2,oo)# and it is concave down on #(-3,2)#

Evelyn Agard · Answer

undefined To determine the concavity of the function $ F(x) = \frac{1}{12}x^4 + \frac{1}{6}x^3 - 3x^2 - 2x + 1 $ and its intervals, you need to find its second derivative. Then, analyze the sign of the second derivative.

1. Find the first derivative $ F'(x) $ by differentiating $ F(x) $ with respect to $ x $.
2. Find the second derivative $ F''(x) $ by differentiating $ F'(x) $ with respect to $ x $.
3. Analyze the sign of $ F''(x) $ to determine concavity:
   - If $ F''(x) > 0 $, the function is concave up in that interval.
   - If $ F''(x) < 0 $, the function is concave down in that interval.

After determining the intervals where the function is concave up or concave down, you can list them accordingly.