Let f(x)=(x^2-8)e^x . How do I determine the inflection points and the intervals on which the given equation is concave up or down?

Question

The equation is f(x)=(x²-8)e^x

Answer 1

You take the second derivative and set it equal to 0 to find the inflection points; this function doesn't have points where the second derivative fails to exist. You should check to see that this is an inflection point. This is necessary but not sufficient if the second derivative exists everywhere. If, on either side of the inflection, the second derivative has opposite sign, then it is an inflection.

Concave up is also referred to as convex; this is where the second derivative is positive. Concave down is where the second derivative is negative. Thus, an inflection point is where the graph switches from being concave up to concave down (or vice-versa, if you are only considering going from left to right).

f(x) = (x^2 - 8)e^x

f'(x) = 2xe^x + (x^2 - 8) e^x

f''(x) = 2e^x + 2xe^x + 2xe^x + (x^2-8)e^x = (x^2 + 4x - 6)e^x

(x^2 + 4x - 6)e^x =0 when (x^2 + 4x - 6) = 0 or x = -2 ± √10

As the second derivative has the same sign as (x^2 + 4x - 6) because e^x is always positive, and x^2 + 4x - 6 is a parabola pointing up, we know that x^2 + 4x - 6 < 0 between its 2 zeroes and positive elsewhere where x^2 + 4x - 6 is not equal to 0. This also tells us that the 2 points where it equals 0 will be inflection points as the sign reverses in both cases at these zeroes.

Thus, we have inflection points at -2 - √10 and -2 + √10.

The function is concave up on (-∞, -2 - √10) and (-2 - √10, ∞) and concave down on (-2 - √10, -2 + √10)

(Note: it is easy to see in a graph the change from concave down to concave up at -2 + √10, approximately 1.16. The other inflection is harder to see in a graph, but it is clear that this switch has to occur, as the function has to be concave up as it goes toward -∞ because the function's derivative is positive but the function has a horizontal asymptotes

To see this inflection, you can separate the graph into [-9, -4] and [-4, 3]

You can also "know" that we will have at least two inflections because the function is continuous but negative on (-√8, √8) and positive on (-∞, -√8) and (√8, ∞), with a horizontal asymptote at y = 0 for negative x.