How do you determine the intervals where #f(x)=x^4-x^2# is concave up or down?

Question

Isaiah Allen · Accepted Answer

Concave up: #(-oo,-1/sqrt(6)), (1/sqrt(6),oo)#
Concave down: #(-1/sqrt(6),1/sqrt(6))# To determine concavity of a function, you must find the second derivative #f''(x)#, determine its critical points (places where #f''(x)=0# or is not defined), and then examine the sign of #f''(x)# in all intervals defined by those critical points. #f(x) = x^4 - x^2# #f'(x) = 4x^3 - 2x# #f''(x) = 12x^2- 2# From examination #f''(x)# is a quadratic function, and therefore there are no places where it is undefined. Thus, our only critical points will come from the zeroes of #f''(x)#: #12x^2-2 = 0# #12x^2 = 2# #x^2 = 1/6 :. x=+-1/sqrt(6)~~+-0.408# Having two zeroes means we have 3 intervals to examine: #(-oo,-1/sqrt(6)), (-1/sqrt(6),1/sqrt(6)), (1/sqrt(6),oo)# Choose a single #x# value inside of each interval and evaluate #f''(x)# at that value. If the result is positive, the function #f(x)# is concave up in that interval; if the result is negative, the function is concave down. For simplicity, choose "easy" values of #x# to evaluate: #f''(-1) = 12(-1)^2-2 = 12-2 = 10 > 0 :. "concave up"# #f''(0) = 12(0)^2-2 = 0-2 = -2 < 0 :. "concave down"# #f''(1) = 12(1)^2-2 = 12-2 = 10 > 0 :. "concave up"# graph{x^4-x^2 [-1.923, 1.923, -0.963, 0.959]}

Charles Anctil · Answer

undefined To determine the intervals where $ f(x) = x^4 - x^2 $ is concave up or down, we need to find the second derivative of $ f(x) $ and then analyze its sign.

First derivative: $ f'(x) = 4x^3 - 2x $

Second derivative: $ f''(x) = 12x^2 - 2 $

To find the intervals where the function is concave up or down, set $ f''(x) = 0 $ and solve for $ x $. Then, test the intervals determined by these critical points with values of $ f''(x) $ in each interval to determine concavity.

Solve $ 12x^2 - 2 = 0 $ to find critical points.

$ 12x^2 - 2 = 0 $
$ 12x^2 = 2 $
$ x^2 = \frac{1}{6} $
$ x = \pm \sqrt{\frac{1}{6}} $

Now, test the intervals created by these critical points with values of $ f''(x) $:

1. Choose a test point in each interval and plug it into $ f''(x) $.
2. Determine the sign of $ f''(x) $ in each interval.

If $ f''(x) > 0 $, the function is concave up in that interval.
If $ f''(x) < 0 $, the function is concave down in that interval.

Finally, state the intervals where the function is concave up or down.