Question

Find the points of local maxima and local minima, if any of the each of the following functions. Find also the local maximum and local minimum values, as the case may be:
(i) $f\left( x \right)=\sin 2x-x$, where $\dfrac{-\pi }{2} < x < \dfrac{\pi }{2}$.
(ii) $f\left( x \right)=\sin x+\dfrac{1}{2}\cos 2x$, where $0 < x < \dfrac{\pi }{2}$.
(iii) $f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x$, where $0 < x < \dfrac{\pi }{2}$.

Answer 1

Hint: We start solving the problem by recalling the procedure of finding the local maxima and local minima. For each of the three problems, we first differentiate the given function and equate that to zero to find the values of x, which may give us the position of local maxima or local minima (extreme points). After finding the values of x, we find the second derivative of the function and substitute in it to check local maxima or minima. After checking we substitute the value of x in the given function to find the value.

Complete step-by-step answer:
(i) According to the problem, we need to find the points of local maxima or local minima (if any) and also the value of local maximum or local minimum for the function $f\left( x \right)=\sin 2x-x$ ---(1), where $\dfrac{-\pi }{2}$$<$$x$$<$$\dfrac{\pi }{2}$.
We know that to find the points of local maxima or local minima of a function $f\left( x \right)$, we need to find the values for the x by solving for ${{f}^{'}}\left( x \right)=0$. We then substitute these values of x in ${{f}^{''}}\left( x \right)$ to check whether it is greater than zero (where local minimum occurs) or less than zero (where local maximum occurs).
Now we differentiate the function $f\left( x \right)=\sin 2x-x$ on both sides with respect to x.
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \sin 2x-x \right)$.
We know that $\dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)-\dfrac{d}{dx}\left( g\left( x \right) \right)$.
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \sin 2x \right)-\dfrac{d}{dx}\left( x \right)\].
We know that $\dfrac{d}{dx}\left( \sin ax \right)=a\cos ax$ and $\dfrac{d}{dx}\left( x \right)=1$.
\[\Rightarrow {{f}^{'}}\left( x \right)=2\cos 2x-1\] ---(2).
Now, we consider ${{f}^{'}}\left( x \right)=0$.
\[\Rightarrow 2\cos 2x-1=0\].
\[\Rightarrow 2\cos 2x=1\].
\[\Rightarrow \cos 2x=\dfrac{1}{2}\].
\[\Rightarrow 2x=\dfrac{\pi }{3}\] or \[2x=\dfrac{-\pi }{3}\].
\[\Rightarrow x=\dfrac{\pi }{6}\] or \[x=\dfrac{-\pi }{6}\] ---(3).
Now, let us find ${{f}^{''}}\left( x \right)$. Let us differentiate \[{{f}^{'}}\left( x \right)=2\cos 2x-1\] on both sides.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( 2\cos 2x-1 \right)$.
We know that $\dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)-\dfrac{d}{dx}\left( g\left( x \right) \right)$.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( 2\cos 2x \right)-\dfrac{d}{dx}\left( 1 \right)$.
We know that $\dfrac{d}{dx}\left( \cos ax \right)=a\sin ax$ and $\dfrac{d}{dx}\left( a \right)=0$.
$\Rightarrow {{f}^{''}}\left( x \right)=-4\sin 2x$.
Let us substitute the values of x we just found in equation (3).
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-4\sin 2\left( \dfrac{\pi }{6} \right)$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-4\sin \left( \dfrac{\pi }{3} \right)$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-4\times \dfrac{\sqrt{3}}{2}$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-2\sqrt{3}$$\left( <0 \right)$. So, we have a local maximum at $x=\dfrac{\pi }{6}$.
Let us find the value of the local maximum of $f\left( x \right)$ at $x=\dfrac{\pi }{6}$ by substituting this in equation (1).
So, we have $f\left( x \right)=\sin 2x-x$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\sin \left( \dfrac{\pi }{3} \right)-\left( \dfrac{\pi }{6} \right)$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}-\left( \dfrac{\pi }{6} \right)$.
Now, we check at $x=\dfrac{-\pi }{6}$.
$\Rightarrow {{f}^{''}}\left( \dfrac{-\pi }{6} \right)=-4\sin 2\left( \dfrac{-\pi }{6} \right)$.
$\Rightarrow {{f}^{''}}\left( \dfrac{-\pi }{6} \right)=-4\sin \left( \dfrac{-\pi }{3} \right)$.
\[\Rightarrow {{f}^{''}}\left( \dfrac{-\pi }{6} \right)=-4\times \left( \dfrac{-\sqrt{3}}{2} \right)\].
$\Rightarrow {{f}^{''}}\left( \dfrac{-\pi }{6} \right)=2\sqrt{3}$$\left( >0 \right)$. So, we have a local minimum at $x=\dfrac{-\pi }{6}$.
Let us find the value of the local minimum of $f\left( x \right)$ at $x=\dfrac{-\pi }{6}$ by substituting this in equation (1).
So, we have $f\left( x \right)=\sin 2x-x$.
$\Rightarrow f\left( \dfrac{-\pi }{6} \right)=\sin \left( \dfrac{-\pi }{3} \right)-\left( \dfrac{-\pi }{6} \right)$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{-\sqrt{3}}{2}+\left( \dfrac{\pi }{6} \right)$.
So, we have a local minimum and local maximum at $x=\dfrac{-\pi }{6}$ and $x=\dfrac{\pi }{6}$. The values of local maximum and local minimum to the function f is $\dfrac{\sqrt{3}}{2}-\left( \dfrac{\pi }{6} \right)$ and $\dfrac{-\sqrt{3}}{2}+\left( \dfrac{\pi }{6} \right)$.

(ii) According to the problem, we need to find the points of local maxima or local minima (if any) and also the value of local maximum or local minimum for the function $f\left( x \right)=\sin x+\dfrac{1}{2}\cos 2x$ ---(4), where $0We know that to find the points of local maxima or local minima of a function $f\left( x \right)$, we need to find the values for the x by solving for ${{f}^{'}}\left( x \right)=0$. We then substitute these values of x in ${{f}^{''}}\left( x \right)$ to check whether it is greater than zero (where local minimum occurs) or less than zero (where local maximum occurs).
Now we differentiate the function $f\left( x \right)=\sin x+\dfrac{1}{2}\cos 2x$ on both side with respect to x.
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \sin x+\dfrac{1}{2}\cos 2x \right)$.
We know that $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)+\dfrac{d}{dx}\left( g\left( x \right) \right)$.
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \sin x \right)+\dfrac{d}{dx}\left( \dfrac{1}{2}\cos 2x \right)\].
We know that $\dfrac{d}{dx}\left( \cos ax \right)=-a\sin ax$ and $\dfrac{d}{dx}\left( \sin x \right)=\cos x$.
\[\Rightarrow {{f}^{'}}\left( x \right)=\cos x+\left( \dfrac{1}{2}\times \left( -2\sin 2x \right) \right)\].
\[\Rightarrow {{f}^{'}}\left( x \right)=\cos x-\sin 2x\] ---(5).
Now, we consider ${{f}^{'}}\left( x \right)=0$.
\[\Rightarrow \cos x-\sin 2x=0\].
We know that $\sin 2x=2\sin x\cos x$.
\[\Rightarrow \cos x\left( 1-2\sin x \right)=0\].
Since $\cos x$ does not becomes 0 in $0$\Rightarrow 1-2\sin x=0$.
$\Rightarrow 2\sin x=1$.
$\Rightarrow \sin x=\dfrac{1}{2}$.
\[\Rightarrow x=\dfrac{\pi }{6}\]---(6).
Now, let us find ${{f}^{''}}\left( x \right)$. Let us differentiate \[{{f}^{'}}\left( x \right)=\cos x-\sin 2x\] on both sides.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( \cos x-\sin 2x \right)$.
We know that $\dfrac{d}{dx}\left( f\left( x \right)-g\left( x \right) \right)=\dfrac{d}{dx}\left( f\left( x \right) \right)-\dfrac{d}{dx}\left( g\left( x \right) \right)$.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( \cos x \right)-\dfrac{d}{dx}\left( \sin 2x \right)$.
We know that $\dfrac{d}{dx}\left( \sin ax \right)=a\cos ax$ and $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$.
$\Rightarrow {{f}^{''}}\left( x \right)=-\sin x-2\cos 2x$.
Let us substitute the value of x we just found in equation (6).
$\Rightarrow {{f}^{''}}\left( x \right)=-\sin x-2\cos 2x$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-\sin \left( \dfrac{\pi }{6} \right)-2\cos 2\left( \dfrac{\pi }{6} \right)$.
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-\sin \left( \dfrac{\pi }{6} \right)-2\cos \left( \dfrac{\pi }{3} \right)$.
\[\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-\dfrac{1}{2}-2\left( \dfrac{\sqrt{3}}{2} \right)\].
\[\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=-\dfrac{1}{2}-\sqrt{3}\]$\left( <0 \right)$. So, we have a local maximum at $x=\dfrac{\pi }{3}$.
Let us find the value of the local maximum of $f\left( x \right)$ at $x=\dfrac{\pi }{6}$ by substituting this in equation (7).
So, we have $f\left( x \right)=\sin x+\dfrac{1}{2}\cos 2x$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\sin \left( \dfrac{\pi }{6} \right)+\dfrac{1}{2}\cos 2\left( \dfrac{\pi }{6} \right)$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}+\dfrac{1}{2}\cos \left( \dfrac{\pi }{3} \right)$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}+\left( \dfrac{1}{2}\times \dfrac{1}{2} \right)$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}+\dfrac{1}{4}$.
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{3}{4}$.
So, we have a local maximum at $x=\dfrac{\pi }{6}$. The values of local maximum and local minimum to the function f is $\dfrac{3}{4}$.

(iii) According to the problem, we need to find the points of local maxima or local minima (if any) and also the value of local maximum or local minimum for the function $f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x$ ---(7), where $0We know that to find the points of local maxima or local minima of a function $f\left( x \right)$, we need to find the values for the x by solving for ${{f}^{'}}\left( x \right)=0$. We then substitute these values of x in ${{f}^{''}}\left( x \right)$ to check whether it is greater than zero (where local minimum occurs) or less than zero (where local maximum occurs).
Now we differentiate the function $f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x$ on both sides with respect to x.
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)$.
\[\Rightarrow {{f}^{'}}\left( x \right)=4{{\sin }^{3}}x\dfrac{d}{dx}\left( \sin x \right)+4{{\cos }^{3}}x\dfrac{d}{dx}\left( \cos x \right)\].
\[\Rightarrow {{f}^{'}}\left( x \right)=4{{\sin }^{3}}x\cos x-4{{\cos }^{3}}x\sin x\]---(8).
Now, we consider ${{f}^{'}}\left( x \right)=0$.
\[\Rightarrow 4{{\sin }^{3}}x\cos x-4{{\cos }^{3}}x\sin x=0\].
\[\Rightarrow 4\sin x\cos x\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)=0\].
\[\Rightarrow 2\sin 2x\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)=0\].
We know that $\sin 2x=0$ at $x=0$, which is not in the interval of x. So, we neglect it.
$\Rightarrow \sin x-\cos x=0$ or $\sin x+\cos x=0$.
$\Rightarrow \sin x=\cos x$ or $\sin x=-\cos x$.
$\Rightarrow \dfrac{\sin x}{\cos x}=1$ or \[\dfrac{\sin x}{\cos x}=-1\].
$\Rightarrow \tan x=1$ or $\tan x=-1$.
\[\Rightarrow x=\dfrac{\pi }{4}\] or \[x=\dfrac{-\pi }{4}\](not in the interval of x. so, we neglect it).
\[\Rightarrow x=\dfrac{\pi }{4}\] ---(9).
Now, let us find ${{f}^{''}}\left( x \right)$. Let us differentiate \[{{f}^{'}}\left( x \right)=4{{\sin }^{3}}x\cos x-4{{\cos }^{3}}x\sin x\] on both sides.
$\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( 4{{\sin }^{3}}x\cos x-4{{\cos }^{3}}x\sin x \right)$.
\[{{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( 4{{\sin }^{3}}x\cos x \right)-\dfrac{d}{dx}\left( 4{{\cos }^{3}}x\sin x \right)\].
\[{{f}^{''}}\left( x \right)=\left( 12{{\sin }^{2}}x\cos x \right)\dfrac{d}{dx}\left( \sin x \right)-4{{\sin }^{4}}x-\left( 12{{\cos }^{2}}x\sin x \right)\dfrac{d}{dx}\left( \cos x \right)-4{{\cos }^{4}}x\].
\[{{f}^{''}}\left( x \right)=\left( 12{{\sin }^{2}}x{{\cos }^{2}}x \right)-\left( 4{{\sin }^{4}}x \right)+\left( 12{{\cos }^{2}}x{{\sin }^{2}}x \right)-4{{\cos }^{4}}x\].
Let us substitute the values of x we just found in equation (3).
\[{{f}^{''}}\left( \dfrac{\pi }{4} \right)=\left( 12{{\sin }^{2}}\left( \dfrac{\pi }{4} \right){{\cos }^{2}}\left( \dfrac{\pi }{4} \right) \right)-\left( 4{{\sin }^{4}}\left( \dfrac{\pi }{4} \right) \right)+\left( 12{{\cos }^{2}}\left( \dfrac{\pi }{4} \right){{\sin }^{2}}\left( \dfrac{\pi }{4} \right) \right)-4{{\cos }^{4}}\left( \dfrac{\pi }{4} \right)\].
\[{{f}^{''}}\left( \dfrac{\pi }{4} \right)=\left( 12{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}} \right)-\left( 4{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}} \right)+\left( 12{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}} \right)-\left( 4{{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}} \right)\].
\[{{f}^{''}}\left( \dfrac{\pi }{4} \right)=\left( 12\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right) \right)-\left( 4\left( \dfrac{1}{4} \right) \right)+\left( 12\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right) \right)-\left( 4\left( \dfrac{1}{4} \right) \right)\].
\[{{f}^{''}}\left( \dfrac{\pi }{4} \right)=3-1+3-1\].
$\Rightarrow {{f}^{''}}\left( \dfrac{\pi }{6} \right)=4$$\left( >0 \right)$. So, we have a local minimum at $x=\dfrac{\pi }{4}$.
Let us find the value of the local minimum of $f\left( x \right)$ at $x=\dfrac{\pi }{4}$ by substituting this in equation (7).
So, we have $f\left( x \right)={{\sin }^{4}}x+{{\cos }^{4}}x$.
$\Rightarrow f\left( \dfrac{\pi }{4} \right)={{\sin }^{4}}\left( \dfrac{\pi }{4} \right)+{{\cos }^{4}}\left( \dfrac{\pi }{4} \right)$.
$\Rightarrow f\left( \dfrac{\pi }{4} \right)=\dfrac{1}{4}+\dfrac{1}{4}$.
$\Rightarrow f\left( \dfrac{\pi }{4} \right)=\dfrac{1}{2}$.
So, we have a local minimum at $x=\dfrac{\pi }{4}$. The values of local minimum to the function f is $\dfrac{1}{2}$.

Note: We should not confuse the local maximum and local minimum after substituting the value of x in the second derivative. We should not calculate mistakes while solving this problem. Since the interval of x is fixed in this problem, we got finite local minima or local maxima. Otherwise, we could have got infinite local minima or local maxima. We can also check these by drawing the plot of the function.