Question

Find the points of local maxima or local minima of the following function, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be.
$f\left( x \right)=\sin x-\cos x,\ 0x < 2\pi$

Answer 1

Hint : To find the points of local maximum or minimum the value of differentiation of function should be zero. So, first of all we will find the differentiation of the equation with respect to x. Then we will equate it with zero and find the value of x which is our value of local minimum or local maximum. Then by substituting the value of x in our function $f\left( x \right)=\sin x-\cos x,\ 0 < x < 2\pi$ . We will get the value of local maxima or local minima.

**Complete step-by-step answer ** :
In question we are asked to find the point of local maxima or local minima of the function $f\left( x \right)=\sin x-\cos x,\ 0 < x < 2\pi$ and also find its local minimum or local maximum values. Now, we know that the to the point at which differentiation of the function is equal to zero is called local maxima or local minima point, which can be seen mathematically as,
$f'\left( x \right)=0$ …………….(i)
Now, here f is function of x, so we will find the value at which value of x will be equal to zero, so the expression of function can be given as,
$f\left( x \right)=\sin x-\cos x$ …………..(ii)
Now, on differentiating the expression with respect to x we will get,
$f'\left( x \right)=\dfrac{d}{dx}\left( \sin x-\cos x \right)$
$\Rightarrow f'\left( x \right)=\dfrac{d}{dx}\left( \sin x \right)-\dfrac{d}{dx}\left( \cos x \right)$
Now, the differentiation of $\sin x$ is $\cos x$ and differentiation of $\cos x$ is $-\sin x$ , so on substituting the values we will get,
$\Rightarrow f'\left( x \right)=\cos x-\left( -\sin x \right)=\cos x+\sin x$
Now, on substituting the value in equation (i) we will get,
$f'\left( x \right)=\cos x+\sin x=0$
$\Rightarrow \sin x=-\cos x$
Now, on dividing $\cos x$ on both the sides we will get,
$\Rightarrow \dfrac{\sin x}{\cos x}=-\dfrac{\cos x}{\cos x}$
Now, we know that $\dfrac{\sin x}{\cos x}=\tan x$ , so we will get,
$\Rightarrow \tan x=-1$
$\therefore x={{\tan }^{-1}}\left( -1 \right)$ ………………(iii)
Now, we know that the value of tan functions are negative in the second and fourth quadrant. So, here for maxima we will take angle in the second quadrant i.e. $180-135=45$ . So, we can say that $\tan \left( {{45}^{\circ }} \right)=\tan \left( 180-135 \right)$ . Now, $\tan \left( 180-135 \right)$ is in the form of $\tan \left( \pi -\theta \right)$ . So, from this we can say that $x={{\tan }^{-1}}\left( 1 \right)=135{}^\circ$ as $x\in \left( 0,\ 2\pi \right)$ . So, on substituting this in expression (iii) we will get,
$x=135$
Now, to find the value of local minima or minima, we will substitute value of x in expression (ii), so we will get,
$f\left( 135 \right)=\sin \left( 135 \right)-\cos \left( 135 \right)$
Now, value of $\sin \left( 135 \right)=\sin \left( 180-45 \right)=\sin \left( 45 \right)$ , same way for cosine, $\cos \left( 135 \right)=\cos \left( 180-45 \right)=-\cos \left( 45 \right)$ . We know that 135 is in the second quadrant where sin functions are positive and cosine functions are negative. So, we can write $\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$ .
On substituting this in equation we will get,
$f\left( 135 \right)=\dfrac{1}{\sqrt{2}}-\left( \dfrac{-1}{\sqrt{2}} \right)=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}$
$\Rightarrow f\left( 135 \right)=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}$
Now, on multiplying and dividing the expression with $\sqrt{2}$ we will get,
$\Rightarrow f\left( 135 \right)=\dfrac{2}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{\sqrt{2}\times \sqrt{2}}$
$\Rightarrow f\left( 135 \right)=\dfrac{2\sqrt{2}}{\sqrt{2}\times \sqrt{2}}=\dfrac{2\sqrt{2}}{2}=\sqrt{2}$ ……………………(iv)
Now, again we will take the fourth quadrant to find local minima. So, here ${{\tan }^{-1}}\left( -1 \right)=-\tan 45{}^\circ =\tan \left( 360-315 \right)$ . Thus, from this we can say that the value of x is 315. So, we will substitute value of x in equation (ii), we get as
$f\left( 315 \right)=\sin \left( 315 \right)-\cos \left( 315 \right)$
On further simplifying, we can write it as
$f\left( 315 \right)=\sin \left( 360-45 \right)-\cos \left( 360-45 \right)$
But we know that in the fourth quadrant angles of sin functions are negative and cosine are positive. So, we get as
$f\left( 315 \right)=\sin \left( -45 \right)-\cos \left( 45 \right)$
$f\left( 315 \right)=-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}$
Now, on multiplying and dividing the expression with $\sqrt{2}$ we will get,
$\Rightarrow f\left( 315 \right)=\dfrac{-2\sqrt{2}}{\sqrt{2}\times \sqrt{2}}$
$\Rightarrow f\left( 315 \right)=\dfrac{-2\sqrt{2}}{2}=-\sqrt{2}$
Thus, local maxima is $\sqrt{2}$ and local minima is $-\sqrt{2}$ .

Note : Remember to find both the angles at which tan function is equal to minus 1. Students sometimes make mistakes and find any one value which will be partially correct. Also, after the equation $\Rightarrow \sin x=-\cos x$ , we can convert it into a cot function also. And by solving in the same way, we will get the same answer as for tan function. So, this can be another approach.