# Is #f(x)=xe^x-x^2# concave or convex at #x=0#?

By signing up, you agree to our Terms of Service and Privacy Policy

To determine whether ( f(x) = xe^x - x^2 ) is concave or convex at ( x = 0 ), we need to examine the second derivative of ( f(x) ) at ( x = 0 ).

( f''(x) = e^x + xe^x - 2 )

Substituting ( x = 0 ) into the second derivative:

( f''(0) = e^0 + 0 \cdot e^0 - 2 = 1 - 2 = -1 )

Since the second derivative is negative at ( x = 0 ), the function is concave at that point.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find points of inflection and determine the intervals of concavity given #y=xe^x#?
- How do you find Find the points of inflection and discuss the concavity of the function #f(x) = 2x^3-3x^2-12x+5#?
- What is the second derivative of #f(x)= ln sqrt(xe^x)#?
- If #f(x)=sinx+cosx#, what are the x-coordinates of the points of inflection of the graph f (x)?
- How do you find local maximum value of f using the first and second derivative tests: #f(x)=x^3-2x+5# on the interval (-2,2)?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7