Question
Asked Oct 23, 2019
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Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) =x/(x2-x+4),  [0, 6]
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Expert Answer

Step 1

Obtain the first derivative of the given function.

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d f(x) dx2-x+4 (x2-x+4)(1)-(x)(2x -1) (x2-x+4) (x-x+4)-(2x -x) (x x+4)* 2 x2 -x+4-2x2+x (x-x+4)* 2 -x2+4 (xx+4) Il

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Step 2

Set f ′(x) = 0 and obtain the critical numbers.

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-x2+4 0 (x2-x+4) -x2+4=0 x 2 Here, the critical numbers are x 2.x -2 Notice that the denominator, 2-x+4 of f (x) is in the form of quadratic equation. -bb-4ac 2a 1 -4(1)(4) 2(1) 115 2 Thus, the roots are not real as the discriminant (15) is less than 0. Since it fails to satisfy the conditions of definition of the critical mumber, there is no critical number for -x+1 Thus, the valid critical numbers are x 2 andx-2 Hence, x 2 lie in the given interval [0.6 and x-2 does not lie in the given interval. Thus, the only valid critical number is x-2

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Step 3

Apply the extreme values of the given interval and the ...

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Substitute x 0 in f(x) 0 f(0) (0)-(0)+4 =0 Substitute x 2 in f(x) 2 f(2) (2)-(2)+4 2 4-2+4 2 6

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