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dem82 [27]
3 years ago
11

A theorem states that if f is continuous on an interval I that contains one local extremum at c and if a local minimum occurs at

c, then f(c) is the absolute minimum of f on I, and if a local maximum occurs at c, then f(c) is the absolute maximum of f on I. Verify that the following function satisfies the conditions of this theorem on its domain. Then, find the location and value of the absolute extremum guaranteed by the theorem.
f(x)= -xe^-x/4 The function f(x) = -xe^-x/4 has an absolute extremum of Q at x=.?
Mathematics
1 answer:
Artist 52 [7]3 years ago
3 0

Answer:

Step-by-step explanation:

To find the extremum of the function, you need to take the first derivative.f'(x) = (-x)'e^{-x/4} + (-x) (e^{-x/4})'

= e^{-x/4}(\frac{x-4}{4} )

This derivative = 0 if and only if x - 4 = 0, hence the extremum is at x = 4

To consider if it is local max or min, you need to consider the act of the function before and after x = 4 by making a table.

-\infty                                        4                                       +\infty

f(x)                 -                       0                +

                     \searrow                                         \nearrow

Hence x =4 is a  local min.

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Look at the image attatched. I have labeled the angles for better explanation.

We know that angle 1 is 54 degrees as is rests on a straight line with its adjacent angle at 126 degrees. This means that angle 1 would be 180 - 126 degrees, or 54 degrees. We also know that angle 1 and 3 must be congruent as the sides of the triangle opposite those angles are congruent as well. This means that both those angles are 54 degrees. Since the sum of all angles in a triangle must equal 180 degrees, we can get that angle 4 is 180 - (2*54) degrees, or 72. Since angle two and angle 4 both lie on a straight line, they must add up to 180 degrees. This means that the value of angle 2 would be 180 - 72, or 108 degrees. Since angle 2 is also equal to x+122, we get the equation:

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