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Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be: f(x) = sinx − cos x, 0 < x < 2π

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प्रश्न

Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:

f(x) = sinx − cos x, 0 < x < 2π

योग
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उत्तर

Given function f(x) = sin x - cos x, 0 < x < 2`pi`

∴ f'(x) = cos x + sin x = cos x (1 + tan x)

If f'(x) = 0 then 1 + tan x = 0

= tan x = - 1

= x = `(3pi)/4, (7pi)/4`

Now f''(x) = `d/dx (cos x + sin x) = - sin x + cos x`

at `x = (3 pi)/4 f' (x) = -sin  (3pi)/4 + cos  (3pi)/4`

`= - (1/sqrt4) - 1/sqrt2`

`= - 2/sqrt2`

`= - sqrt2`              ...(negative)

∴ f(x) is maximum at `x = (3pi)/4`.

and the maximum value of f(x)

`f((3pi)/4)= sin  (3pi)/4 - cos  (3pi)/4`

`= 1/sqrt2 - (- 1/sqrt2)`

`= 2/sqrt2`

`= sqrt2`

Again, at `x = (7pi)/4  f' (x) = -sin  (7pi)/4 + cos  (7pi)/4`

`= - ((-1)/sqrt2) + 1/sqrt2`

`= 2/sqrt2`

`= sqrt2`            ... (positive)

∴ f(x) is minimum at `x = (7 pi)/4`.

and the minimum value of f(x)

`= f ((7pi)/4) = sin  ((7pi)/4) - cos  ((7pi)/4)`

`= - 1/sqrt2 - 1/sqrt2`

`= - sqrt2`

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अध्याय 6: Application of Derivatives - Exercise 6.5 [पृष्ठ २३२]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.5 | Q 4 | पृष्ठ २३२

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