Find the absolute maximum and absolute minimum values of f on the given interval.

Question

f(x) = xe^−x2/72,    [−4, 12]

Answer 1

f(x) = xe^{-(x^2)/72}

f(x) = xe^{-(x^2/72)}

f'(x) = e^{-(x^2)/72} + x(-x/36)e^{-(x^2)/72} = e^{-(x^2)/72} (1 - x²/36) = 0

1 - x²/36 = 0 So, x = 6 or -6.

6 is the only critical value in the interval [-4, 12]

To find the absolute extrema of f on the interval [-4,12], evaluate f at the endpoints and at the critical values in the interval.

f(6) = 6e^-0.5 ≈ 3.64

f(12) = 12e^-2 ≈ 1.62

f(-4) = -4e^-0.222 ≈ -3.2

Answer 2

f(x) = xe^{(-x^2)/72}

f'(x) = xe^{(-x^2)/72}(-2x/72) + (1)e^{(-x^2)/72} [product rule, chain rule]

Set f'(x) = 0

xe^{(-x^2)/72}(-2x/72) + (1)e^{(-x^2)/72} = 0

e^{(-x^2)/72} (-2x²/72 + 1) = 0

e^{(-x^2)/72} ≠ 0

-2x²/72 + 1 = 0

Solve for x.

2x² = 72

x² = 36

x = -6 or 6

-6 is not in the interval [-4, 12]; x = 6 is in the interval [-4, 12]

Use the sign test for f'(x).

f'(x) > 0 for x < 6 and f'(x) > 0 for x > 6

f'(x) changes sign from positive to negative at x = 6; therefore, x = 6 is a maximum.

Check:

f(-4), f(6), f(12)

f(-4) = -3.2

f(6) = 3.6

f(12) = 1.6

For (x) = xe^{(-x^2)/72} in the interval [-4, 12]

The absolute maximum is at x = 6.

The absolute minimum is at x = -4

Answer 3

Graph this!

The absolute minimum occurs at x=-4. Find by graphing.

The absolute maximum occurs at x=6 . Use product rule to find derivative and set it=0.