True/false:
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?
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 )
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.
To determine the concavity of a function, you need to consider the sign of its second derivative. If the second derivative is positive, the function is concave up, and if it is negative, the function is concave down.
In this case, you correctly found that the second derivative of the function x^2 - 3 is 2. Since the second derivative is positive (greater than 0), the function is concave up.
Now, to determine the concavity on a specific interval, such as (-1, 1), you need to evaluate the second derivative at various points within that interval.
To do this, you can choose any value between -1 and 1, for example, x = 0. Substitute this value into the second derivative equation:
f''(0) = 2
Since f''(0) is positive, the function x^2 - 3 is concave up on the interval (-1, 1).