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

f(x) = xe^−x^2/32, [−2, 8]

The question is missing parts. Here is the complete question.

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

[tex]f(x)=xe^{-\frac{x^{2}}{32} }[/tex] , [ -2,8]

Answer: Absolute maximum: f(4) = 2.42;

              Absolute minimum: f(-2) = -1.76;

Step-by-step explanation: Some functions have absolute extrema: maxima and/or minima.

Absolute maximum is a point where the function has its greatest possible value.

Absolute minimum is a point where the function has its least possible value.

The method for finding absolute extrema points is

1) Derivate the function;

2) Find the values of x that makes f'(x) = 0;

3) Using the interval boundary values and the x found above, determine the function value of each of those points;

4) The highest value is maximum, while the lowest value is minimum;

For the function given, absolute maximum and minimum points are:

[tex]f(x)=xe^{-\frac{x^{2}}{32} }[/tex]

Using the product rule, first derivative will be:

[tex]f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )[/tex]

[tex]f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )[/tex] = 0

[tex]1-\frac{x^{2}}{16}=0[/tex]

[tex]\frac{x^{2}}{16}=1[/tex]

[tex]x^{2}=16[/tex]

x = ±4

x can't be -4 because it is not in the interval [-2,8].

[tex]f(-2)=-2e^{-\frac{(-2)^{2}}{32} }=-1.76[/tex]

[tex]f(4)=4e^{-\frac{4^{2}}{32} }=2.42[/tex]

[tex]f(8)=8e^{-\frac{8^{2}}{32} }=1.08[/tex]

Analysing each f(x), we noted when x = -2, f(-2) is minimum and when x = 4, f(4) is maximum.

Therefore, absolute maximum is f(4) = 2.42 and

absolute minimum is f(-2) = -1.76

The absolute maximum and absolute minimum values of f(x) are 2.426 and -1.765, respectively.

The function is given as:

[tex]\mathbf{f(x) = xe^{-\frac{x^2}{32}}}[/tex]

Start by differentiating the function

[tex]\mathbf{f'(x) = e^{-\frac{x^2}{32}} - \frac{x^2}{16}e^{-\frac{x^2}{32}}}[/tex]

Set to 0

[tex]\mathbf{ e^{-\frac{x^2}{32}} - \frac{x^2}{16}e^{-\frac{x^2}{32}} = 0}[/tex]

Rewrite as:

[tex]\mathbf{ \frac{x^2}{16}e^{-\frac{x^2}{32}} = e^{-\frac{x^2}{32}}}[/tex]

Cancel out common factors

[tex]\mathbf{ \frac{x^2}{16}=1}[/tex]

Multiply both sides by 16

[tex]\mathbf{ x^2=16}[/tex]

Take square roots of both sides

[tex]\mathbf{ x=\pm 4}[/tex]

The interval is given as: [-2,8]

This means that the value of x = -4, is out of the interval.

So, we have the following critical points:

x = (-2, 4, 8)

Calculate f(x) at the these points

[tex]\mathbf{f(-2) = -2 \times e^{-\frac{(-2)^2}{32}}}[/tex]

[tex]\mathbf{f(-2) = -2 \times e^{-\frac{4}{32}}}[/tex]

[tex]\mathbf{f(-2) = -1.765}[/tex]

[tex]\mathbf{f(4) = 4 \times e^{-\frac{(4)^2}{32}}}[/tex]

[tex]\mathbf{f(4) = 4 \times e^{-\frac{16}{32}}}[/tex]

[tex]\mathbf{f(4) = 2.426}[/tex]

[tex]\mathbf{f(8) = 8 \times e^{-\frac{(8)^2}{32}}}[/tex]

[tex]\mathbf{f(8) = 8 \times e^{-\frac{64}{32}}}[/tex]

[tex]\mathbf{f(8) = 1.083}[/tex]

By comparison, the absolute maximum and absolute minimum values of f(x) are 2.426 and -1.765, respectively.

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