Answer to Question #251109 in Real Analysis for Sourav Mondal

Question #251109
Check whether the function f, given by
f(x) = cos x - sin x has an extreme value in the interval (-à ƒ ƒ  €/2,à ƒ ƒ  €/2).
1
Expert's answer
2021-10-14T18:15:08-0400

f(x) = cos x - sin x

Differentiating with respect to x

f'(x) = - sin x - cos x

f''(x) = - cos x + sin x

For extreme values of f(x) ,

f'(x) = 0

=> - sin x - cos x = 0

=> sin x = - cos x

=> tan x = -1

=> tan x = tan (π4)-\frac{π}{4})

=> x = nπ π4-\frac{π}{4} Where n is an integer

Since x (π2,π2)\in(-\frac{π}{2}, \frac{π}{2}) the only value of n is 0.

So x = π4-\frac{π}{4}

f''(π4)=cos(π4)+sin(π4)-\frac{π}{4}) = -cos(-\frac{π}{4})+sin(-\frac{π}{4})

= cos(π4)sin(π4)-cos(\frac{π}{4})-sin(\frac{π}{4})

= 1212-\frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}}

= 22-\frac{2}{\sqrt{2}}

= 2-\sqrt{2}

So f''(π4)-\frac{π}{4}) is negative.

Therefore f(x) has a maximum value at x = π4-\frac{π}{4} and maximum value is f(π4)=cos(π4)sin(π4)f(-\frac{π}{4}) = cos(-\frac{π}{4})-sin(-\frac{π}{4})

= 12+12\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}}

=2\sqrt{2}

Answer:

Yes , there is a extreme value of f(x) = cos x - sin x at x = π4-\frac{π}{4} and it is a maximum value equating 2\sqrt{2}


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