Question

on what interval is the following curve concave downward? (enter your answer using interval notation.) y = x t2 t2 t 9 dt 0

Answer 1

The curve
`\(y = \int_(0)^(x) t^2(t^2 - 9) \,dt\)` is concave downward on the intervals
`\((-\infty, -3)\)` and
`\((3, \infty)\)`.

Step-by-step explanation:

To determine the intervals where the given curve is concave downward, we need to analyze the second derivative. First, find the first derivative of the function
`\(y\)`:

`\[y' = x^2 - 9.\]`

Now, find the second derivative:

`\[y'' = 2x.\]`

For concavity, we are interested in the sign of the second derivative. The second derivative
`\(2x\)` changes sign at
`\(x = 0\)`. When
`\(x < 0\), \(2x\)` is negative, indicating concavity downward. Therefore, the curve is concave downward on the interval
`\((-\infty, 0)\)`. Similarly, when
`\(x > 0\), \(2x\)` is positive, so the curve is concave downward on the interval
`\((0, \infty)\)`.

However, we need to consider the points where the concavity changes. At
`\(x = -3\)` and
`\(x = 3\)`, the sign of the second derivative changes. Therefore, we exclude these points from the intervals. Combining these results, we get that the curve is concave downward on the intervals
`\((-\infty, -3)\)` and
`\((3, \infty)\).`

Answer 2

The function y = integral from 0 to x of t^2 + 9 dt is concave downward on the interval (-∞, 0), as the second derivative 2x is negative for any negative value of x.

Step-by-step explanation:

To determine on what interval the curve given by the integral function y = ∫_{0}^{x} t^2 + 9 dt is concave downward, we need to find the second derivative of y with respect to x and then analyze its sign. The first derivative of y, which we will call y', is simply the integrand evaluated at the upper limit of integration due to the Fundamental Theorem of Calculus, so y' = x^2 + 9. Now, taking the derivative of y' with respect to x to find the second derivative, denoted as y'', we get y'' = 2x.

A function is concave downward wherever the second derivative is negative. As y'' = 2x, this expression is negative when x is negative. Therefore, the interval on which the curve is concave downward is (-∞, 0).