Continuity and Differentiability Test - 17

The function $$\displaystyle f(x)=x^{3}-4x$$ is a polynomial and so it is continuous and differentiable at all x$$\displaystyle \epsilon $$ R.
In particular it is continuous in the closed interval $$[-2,2]$$ .
Also $$f(-2)]=0=f(2)$$. Thus , $$f(x)$$ satisfies all three conditions of Rolle's theorem in $$(-2,2)$$.
Therefore , there must exist at least one real number $$'x'$$ in the
open interval $$(-2,2)$$ for which $$f'(x)=0$$
Also $$\displaystyle f'(x)= 3x^{2}-4$$
Now $$\displaystyle f^{'}(x)=0$$ gives $$3x^{2}-4=0$$ or $$x=\pm \dfrac{2}{\sqrt{3}}$$ which is also known as stationary point.
Both these value lie in the open interval $$(-2,2)$$ and thus the conclusion of Rolle's theorem is verified.

It can be easily seen that $$f(x)=x^{2}-3x+2$$ is continuous as differentiable on R (being a polynomial) $$\Rightarrow f(x)$$ is continous in (1,2) and differentiable in [1,2]. Also, we have

$$f(1)=f(2)=0$$.

Let  $$y = \displaystyle (\tan x)^{\log x}$$
$$\log y = \log x .\log \tan x$$
Differentiating both side w.r.t $$x$$
$$\cfrac{1}{y}.\cfrac{dy}{dx} = \cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x)$$
$$\Rightarrow \cfrac{dy}{dx} =  (\tan x)^{\log x} \left[\cfrac{\log \tan x}{x}+\cfrac{\log x}{\tan x}(\sec^2x) \right]$$

The function $$f\left( x \right)=\sin { x } $$ is continuous and differentiable for all $$x\in R$$.

We have $$f\left( a \right) =f\left( 0 \right) =0$$ and $$\displaystyle f\left( b \right) =f\left( \frac { 1 }{ 2 }  \right) =\frac { 3 }{ 8 } $$
$$\displaystyle \therefore \frac { f\left( b \right) -f\left( a \right)  }{ b-a } =\frac { \frac { 3 }{ 8 } -0 }{ \frac { 1 }{ 2 } -0 } =\frac { 3 }{ 4 } $$
Now 
$$f\left( x \right) ={ x }^{ 3 }-3{ x }^{ 2 }+2x\\ \Rightarrow f'\left( x \right) ={ 3x }^{ 2 }-6x+2\\ \Rightarrow f'\left( c \right) ={ 3c }^{ 2 }-6c+2$$
Putting all these value in lagrange's mean value theorem
$$\displaystyle \frac { f\left( b \right) -f\left( a \right)  }{ b-a } =f'\left( c \right) ,\left( a<c,b \right) $$
We get $$\displaystyle \frac { 3 }{ 4 } ={ 3c }^{ 2 }-6c+2\Rightarrow c=1\pm \frac { \sqrt { 21 }  }{ 6 } $$
Hence $$\displaystyle c=\frac { 1-\sqrt { 21 }  }{ 6 } $$ lies in the open interval $$\displaystyle \left( 0,\frac { 1 }{ 2 }  \right) $$
Therefore it is the required value

$$\displaystyle f\left( x \right)={ x }^{ \frac { 2 }{ 3 }  }\Rightarrow f'\left( x \right) =\frac { 2 }{ x } { x }^{ -\frac { 1 }{ 3 }  }$$
$$\displaystyle\therefore \lim _{ x\rightarrow 0 }{ f'\left( x \right) } =\lim _{ x\rightarrow 0 }{ \frac { 2 }{ 3 } { \left( 0+h \right)  }^{ -\frac { 1 }{ 3 }  } } =\infty$$
$$\displaystyle Rf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right)  }{ h }  \right\}  } =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { h }^{ \frac { 2 }{ 3 }  }-1 }{ h }  \right\}  } =+\infty$$
$$\displaystyle Lf'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { f\left( 0+h \right) -f\left( 0 \right)  }{ -h }  \right\}  } =f'\left( 0 \right) =\lim _{ h\rightarrow 0 }{ \left\{ \frac { { \left( -h \right)  }^{ \frac { 2 }{ 3 }  } }{ h }  \right\}  } =-\infty$$
$$\displaystyle \therefore Lf'\left( 0 \right) \neq Rf'\left( 0 \right)$$
$$\displaystyle\therefore f'\left( 0 \right)$$ does not exist showing that $$f'(x)$$ does not exists in the open interval $$(-1,1)$$
Hence, Rolle's Theorem is not applicable
although $$f(-1)=f(1)=1$$ and $$f(x)$$ is continuous in the closed interval $$(-1,1)$$

$$\dfrac{dx}{dt}=\dfrac{d}{dt}\left(2\cos \left(t\right)-\cos \left(2t\right)\right)$$

$$f(0-) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos -h)=\lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1$$, Since $$\cos(-x) = \cos x$$
and $$f(0+) = \displaystyle \lim_{h\to 0}\sin ^{-1}(\cos h) = \sin^{-1}1$$
Clearly at $$x=0$$, L.H.L $$=$$  R.H.L $$=f(0) \Rightarrow f(x)$$ is continuous at $$x =0$$
Also  $$\displaystyle {y}'=-\frac{\sin x}{\sqrt{1-\cos ^{2}x}}=-\frac{\sin

x}{\sqrt{\sin ^{2}x}}=-\frac{\sin x}{\left | \sin x \right |}$$
So the function is not differentiable at the points where $$\sin x=0$$,
that is, for $$x=k\pi \left ( k\in I \right )$$. In particular, $$x=0$$.