How do you determine intervals on which the function is concave up or down and find the points of inflection for #y=(x^2-7)e^x#?

Question

Paisley Johnson · Accepted Answer

Have a look:

And you can even check the graph:

graph{(x^2-7)e^x [-16.01, 16.04, -7.98, 8.05]}

Emily Ammerman · Answer

undefined To determine intervals of concavity and points of inflection for the function $ y = (x^2 - 7)e^x $, follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for critical points.
3. Use the second derivative test to determine intervals of concavity and points of inflection.

Let's go through these steps:

1. First derivative of $ y = (x^2 - 7)e^x $:
$$ \frac{dy}{dx} = (2x - 7)e^x + (x^2 - 7)e^x $$
$$ \frac{dy}{dx} = (x^2 - 5x - 7)e^x $$

2. Second derivative of $ y = (x^2 - 7)e^x $:
$$ \frac{d^2y}{dx^2} = (2x - 5)(x^2 - 7)e^x + (x^2 - 5x - 7)e^x $$
$$ \frac{d^2y}{dx^2} = (x^3 - 10x + 35)e^x $$

3. Set the second derivative equal to zero:
$$ (x^3 - 10x + 35)e^x = 0 $$

The exponential term $ e^x $ is never zero, so we only need to solve for $ x^3 - 10x + 35 = 0 $. However, this cubic equation doesn't have real roots, which means there are no critical points to consider.

Now, to determine the intervals of concavity and points of inflection, you need to consider the sign of the second derivative in different intervals. Since the second derivative is always positive for all real numbers $ x $ (since the cubic term dominates for large $ x $ values and is always positive), the function is concave up for all $ x $ and has no points of inflection.

In summary:
- The function $ y = (x^2 - 7)e^x $ is concave up for all real numbers $ x $.
- There are no points of inflection for this function.