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

1 Answer
Oct 1, 2017

Concave up: (-oo,-1/sqrt(6)), (1/sqrt(6),oo)
Concave down: (-1/sqrt(6),1/sqrt(6))

Explanation:

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]}