Question

Find the local maxima and local minima of the function $f(x) = \sin x - \cos x,0 < x < 2\pi$ . Also find the local maximum and local minimum values.

Answer 1

Before solving the question, you should know what maxima and minima of a function mean. Maxima and minima of any function are collectively known as extrema. They are the maximum and minimum values of a function respectively. The local maxima and the local minima are the maximum and minimum values of the function in a given range. To find the local extrema, you have to find critical points of the function f(x)by differentiating it and then equating it to zero.

Step-wise-solution:
Give, the function $f(x) = \sin x - \cos x,0 < x < 2\pi$
To find local maxima and local minima of the function f(x), and also the maximum and minimum values of the function f(x) in the given range.
We begin the solution by differentiating f(x) and then equating it to zero, i.e.
$\dfrac{{df(x)}}{{dx}} = 0$
$\Rightarrow \cos x + \sin x = 0$
$\Rightarrow \cos x = - \sin x$
$\Rightarrow \tan x = - 1$
For$\tan x = - 1$, $x = \dfrac{{3\pi }}{4}$ and $x = \dfrac{{7\pi }}{4}$
We know for local maxima, $f''(x)$ has to be negative, and for local minima, $f''(x)$ has to be positive.

Putting $x = \dfrac{{3\pi }}{4}$ , we have $- \sqrt 2$as the result, which is a negative number.
Therefore, at $x = \dfrac{{3\pi }}{4}$ , f(x) has a local maximum.
Putting $x = \dfrac{{7\pi }}{4}$ , we have $\sqrt 2$ as the result, which is a positive number.
Therefore, at $x = \dfrac{{7\pi }}{4}$ , f(x) has a local minimum.
The local minimum value is $- \sqrt 2$and the local minimum value is $\sqrt 2$

Note:
To solve such questions, you should have the knowledge of trigonometric functions and also about maxima and minima. You will find many basic questions in the NCERT book to start with and then proceed to competitive books to enhance your knowledge.