Asked by James
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?
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?
Answers
Answered by
Bosnian
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 )
In this case f"(x)= 2
So x^2-3 is concave up for all vales of x, including interval ( -1, 1 )
Answered by
Steve
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.
Don't forget your general knowledge of functions in the search for a calculus solution.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.