Please show all work so that i may understand your answer, thank you for your time! Find the absolute maximum and absolute minimum of f(r) on the given intervalf(x) =[-1, 2]%3Dx2 + 1'f (x) = sin(x) + cos(x),[0,

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Description: Please show all work so that i may understand your answer, thank you for your time! Find the absolute maximum and absolute minimum of f(r) on the given intervalf(x) =[-1, 2]%3Dx2 + 1'f (x) = sin(x) + cos(x),[0,

1) fx=x2x2+1, -1,2 Differentiate f(x) with respect to x…

Answered: Find the absolute maximum and absolute…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Please show all work so that i may understand your answer, thank you for your time!

Find the absolute maximum and absolute minimum of f(r) on the given interval
f(x) =
[-1, 2]
%3D
x2 + 1'
f (x) = sin(x) + cos(x),
[0,

Expert Solution

Step 1

1) $f (x) = \frac{x^{2}}{x^{2} + 1}, [- 1, 2]$

Differentiate f(x) with respect to x

$\begin{array}{rcl} f' (x) & = & \frac{x^{2} + 1 (x^{2})' - x^{2} (x^{2} + 1)'}{x^{2} + 1} \\ = & \frac{x^{2} + 1 (2 x) - x^{2} (2 x)}{x^{2} + 1} \\ = & \frac{2 x^{3} + 2 x - 2 x^{3}}{x^{2} + 1} \\ = & \frac{2 x}{x^{2} + 1} \end{array}$

Step 2

To find the critical values of the given function, equate f'(x)=0

$\begin{array}{rcl} f' (x) & = & 0 \\ \frac{2 x}{x^{2} + 1} & = & 0 \\ 2 x & = & 0 \\ x & = & 0 \end{array}$

The critical point is x=0

Plug the critical point & the endpoints of the interval in f(x) to get absolute minimum and absolute maximum.

$f (0) = \frac{0}{0 + 1} = 0 f (- 1) = \frac{1}{1 + 1} = \frac{1}{2} = 0.5 f (2) = \frac{4}{4 + 1} = \frac{4}{5} = 0.8$

Therefore, at x=0, f(x) is absolutely minimum & at x=2 f(x) is absolutely maximum.

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