Question

Find the absolute minimum and absolute maximum values of f(x)=xe^−x²​/18 on the interval [−2,6].

Answer 1

The absolute minimum value of
`(f(x) = (x e^(-x^2))/(18)\)` on the interval
`\([-2,6]\)` is approximately -0.0039, and the absolute maximum value is approximately 0.0674.

Step-by-step explanation:

To find the critical points of
`\(f(x)\)`, we first take the derivative \(f'(x)\) and set it equal to zero. After solving for
`(x\))`, we obtain critical points. By evaluating
`\(f(x)\)` at these points and the endpoints of the interval, we determine the absolute minimum and maximum values.

Upon solving, we find that the critical points are
`\(x \approx -0.821\)` and
`\(x \approx 0.821\)`. Evaluating
`\(f(x)\)` at these points and the interval endpoints, we get the values -0.0039, 0.0674, -0.0016, and 0.0597, respectively. Therefore, the absolute minimum is approximately -0.0039 at
`\(x \approx -0.821\)`, and the absolute maximum is approximately 0.0674 at
`\(x \approx 0.821\)`.

By analyzing the behavior of
`\(f(x)\)` as (x) approaches the interval endpoints, we confirm these as the absolute minimum and maximum values. This is because (f(x)) tends towards negative infinity as (x) approaches -2 and towards zero as \(x\) approaches 6. Therefore, the overall absolute minimum and maximum occur at the critical points mentioned earlier.