Question

i) The function f(x) = cos x - sin x has maximum or minimum value at
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Answer 1

$\large\underline\blue{\bold{Given \:  Question   }}$

f(x) = cos x - sin x

$\large\underline\blue{\bold{To \:  Find   }}$

Local Maximum and Local Minimum

$\large\underline\blue{\bold{Theory }}$

How to Find Maximum and Minimum Points Using Differentiation ?

The value of the function at a maximum point is called the maximum value of the function and the value of the function at a minimum point is called the minimum value of the function.

• Differentiate the given function.

• let f'(x)  =  0 and find critical numbers

• Then find the second derivative f''(x).

• Apply those critical numbers in the second derivative.

• The function f (x) is maximum when f''(x) < 0

• The function f (x) is minimum when f''(x) > 0

• To find the maximum and minimum value we need to apply those x values in the given function.

$\large\underline\blue{\bold{Solution :- }}$

$\tt \: f(x) = cos x - sin x - - (i)$

On differentiating both sides w. r. t. x, we get

$:\implies \tt \: f'(x) \: = \: - \: sinx \: - \: cosx - - (ii)$

For maximum or minimum value,

$:\implies \boxed{ \pink{\tt \: f'(x) = 0}}$

$:\implies \tt \: - sinx \: - cosx = 0$

$:\implies \tt \: - cosx = sinx$

$:\implies \tt \: \dfrac{sinx}{cosx} = - 1$

$:\implies \tt \: tanx \: = \: - \: 1$

As we know, tanx < 0 in second and fourth quadrant and tanx = 1 at 45°.

$:\implies \tt \: x = \pi \: - \: \dfrac{\pi}{4} \: or \: x \: = \: 2\pi \: - \dfrac{\pi}{4}$

$:\implies \boxed{ \pink{ \tt \: x \: = \: \: \dfrac{3\pi}{4} \: or \: x \: = \: \: \dfrac{7\pi}{4} }}$

Now, again differentiating (i) both sides w. r. t. x, we get

$:\implies \tt \: f''(x) = sinx \: - \: cosx$

$:\implies \tt \: At \: x \: = \: \dfrac{3\pi}{4}$

$:\implies \tt \: f''(\: \dfrac{3\pi}{4} ) = sin\: \dfrac{3\pi}{4} - cos\: \dfrac{3\pi}{4}$

$:\implies \tt \: f''(\: \dfrac{3\pi}{4} ) =sin(\pi \: - \: \dfrac{\pi}{4} ) - cos(\pi \: - \: \dfrac{\pi}{4} )$

$:\implies \tt \: f''(\: \dfrac{3\pi}{4} ) =sin\: \dfrac{\pi}{4} + cos\: \dfrac{\pi}{4}$

$:\implies \tt \: f''(\: \dfrac{3\pi}{4} ) =\dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} }$

$:\implies \tt \: f''(\: \dfrac{3\pi}{4} ) =\dfrac{2}{ \sqrt{2} }$

$:\implies \tt \: f''(\: \dfrac{3\pi}{4} ) = \sqrt{2}$

$:\implies \tt \: f''(\: \dfrac{3\pi}{4} ) > 0$

$:\implies \boxed{ \pink{\tt \: f(x) \: is \: minimum \: at \: x = \: \dfrac{3\pi}{4} }}$

$:\implies \tt \: minimum \: value \: is \: \red{ \bf \: \: f(\: \dfrac{3\pi}{4} ) }$

$:\implies \tt \: f(\: \dfrac{3\pi}{4} ) =cos\: \dfrac{3\pi}{4} - sin\: \dfrac{3\pi}{4}$

$:\implies \tt \: f(\: \dfrac{3\pi}{4} ) =cos\: (\pi \: - \dfrac{\pi}{4}) - sin\:(\pi \: - \dfrac{\pi}{4} )$

$:\implies \tt \: f(\: \dfrac{3\pi}{4} ) = - \: cos\: \dfrac{\pi}{4} - sin\: \dfrac{\pi}{4}$

$:\implies \tt \: f(\: \dfrac{3\pi}{4} ) = \: - \dfrac{1}{ \sqrt{2} } - \dfrac{1}{ \sqrt{2} }$

$:\implies \tt \: f(\: \dfrac{3\pi}{4} ) = - \: \dfrac{2}{ \sqrt{2} }$

$:\implies \boxed{ \pink{ \tt \: f(\: \dfrac{3\pi}{4} ) = - \sqrt{2} }}$

$\bf \red{now}$

$:\implies \tt \: At \: x \: = \: \dfrac{7\pi}{4}$

$:\implies \tt \: f''(\: \dfrac{7\pi}{4} ) =sin\: \dfrac{7\pi}{4} - cos\: \dfrac{7\pi}{4}$

$:\implies \tt \: f''(\: \dfrac{7\pi}{4} ) =sin(2\pi \: - \: \dfrac{\pi}{4} ) - cos(2\pi \: - \: \dfrac{\pi}{4} )$

$:\implies \tt \: f''(\: \dfrac{7\pi}{4} ) = - sin\: \dfrac{\pi}{4} - cos\: \dfrac{\pi}{4}$

$:\implies \tt \: f''(\: \dfrac{7\pi}{4} ) = - \dfrac{1}{ \sqrt{2} } - \dfrac{1}{ \sqrt{2} }$

$:\implies \tt \: f''(\: \dfrac{7\pi}{4} ) = - \dfrac{2}{ \sqrt{2} }$

$:\implies \tt \: f''(\: \dfrac{7\pi}{4} ) = - \sqrt{2}$

$:\implies \tt \: f''(\: \dfrac{7\pi}{4} ) < 0$

$:\implies \boxed{ \pink{ \tt \: f(x) \: is \: maximum \: at \: x \: = \: \dfrac{7\pi}{4} }}$

$:\implies \tt \: maximum \: value \: is \: \red{ \bf \: \: f(\: \dfrac{7\pi}{4} ) }$

$:\implies \tt \: f(\: \dfrac{7\pi}{4} ) =cos\: \dfrac{7\pi}{4} - sin\: \dfrac{7\pi}{4}$

$:\implies \tt \: f(\: \dfrac{7\pi}{4} ) =cos\: (2\pi \: - \dfrac{\pi}{4}) - sin\:(2\pi \: - \dfrac{\pi}{4} )$

$:\implies \tt \: f(\: \dfrac{7\pi}{4} ) = \: cos\: \dfrac{\pi}{4} + sin\: \dfrac{\pi}{4}$

$:\implies \tt \: f(\: \dfrac{7\pi}{4} ) = \: \dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} }$

$:\implies \tt \: f(\: \dfrac{7\pi}{4} ) = \: \dfrac{2}{ \sqrt{2} }$

$:\implies \boxed { \pink{ \tt \: f(\: \dfrac{7\pi}{4} ) = \: \sqrt{2} }}$

Hence,

$\begin{gathered}\begin{gathered}\bf cosx - sinx \: has \: \begin{cases} &\sf{minimum \: value \: at \: x \: = \: \dfrac{3\pi}{4} } \\ &\sf{maximum \: value \: at \: x \: = \: \dfrac{7\pi}{4} } \end{cases}\end{gathered}\end{gathered}$