True/false:

Question

x^2-3 is concave up on the interval (-1,1)

The second derivative of x^2-3 is just 2. How do I determine concavity of a function when I can't set it equal to 0? And then how do I answer this question when it's asking for a specific interval?

Answer 1

f(x) is concave up on an interval if f"(x) > 0

In this case f"(x)= 2

So x^2-3 is concave up for all vales of x, including interval ( -1, 1 )

Answer 2

you only set f"=0 to find points of inflection. A parabola has no such points.

Don't forget your general knowledge of functions in the search for a calculus solution.