Answer to Question #251109 in Real Analysis for Sourav Mondal

Question

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).

Answer 1

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 ($-\frac{π}{4})$

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

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

So x = $-\frac{π}{4}$

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

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

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

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

= $-\sqrt{2}$

So f''($-\frac{π}{4})$ is negative.

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

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

=$\sqrt{2}$

Answer:

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