Question

If $f(x)=\sqrt{\left| x-1 \right|}$ and$g(x)=\sin x$ then calculate$(fog)x$ and $(gof)x$ discuss differentiability of $(gof)x$ at x=1.

Answer 1

Hint: The given problem is related to composite functions and their differentiability. Use the formula given below and check the differentiability by removing the modulus sign.
$(gof)x=g[f(x)]$

Complete step-by-step solution -
We will write the given first,
$f(x)=\sqrt{\left| x-1 \right|}$And$g(x)=\sin x$ …………………………………….. (1)
Now, we will find $(fog)x$ and $(gof)x$ separately,
(1). $(fog)x=f[g(x)]$
As, $f(x)=\sqrt{\left| x-1 \right|}$
To find $f[g(x)]$ put $x=g(x)$i.e. put, $x=\sin x$
$\therefore (fog)x=\sqrt{\left| \sin x-1 \right|}$
(2). $(gof)x=g[f(x)]$
As, $g(x)=\sin x$
To find$g[f(x)]$ put$x=f(x)$ i.e. put, $x=\sqrt{\left| x-1 \right|}$
$\therefore (gof)x=\sin \left( \sqrt{\left| x-1 \right|} \right)$……………………………………… (2)
Now to check the differentiability of function given in equation (2) we should define it by removing its modulus first,

Modulus functions can be defined as shown below,
If $f(x)=\left| x \right|$ then,
$f(x)=-x$ For $x<0$
$f(x)=x$ For $x\ge 0$
Similarly we can define $(gof)x$ with the help of above definition,
$(gof)x=\sin \left( \sqrt{-(x-1}) \right)$ For, $x-1<0$
$(gof)x=\sin \left( \sqrt{x-1} \right)$ For, $x-1\ge 0$
If we have to simplify the limits then we should add it by 1 on both sides,
$(gof)x=\sin \left( \sqrt{-(x-1}) \right)$ For, $x-1+1<0+1$
$(gof)x=\sin \left( \sqrt{x-1} \right)$ For, $x-1+1\ge 0+1$
Now by further algebraic simplifications we can write above equations as,
$(gof){{x}^{-}}=\sin \left( \sqrt{-x+1} \right)$ For, $x<1$
$(gof){{x}^{+}}=\sin \left( \sqrt{x-1} \right)$ For, $x\ge 1$
Now as asked in problem we have to check the differentiability of $(gof)x$ at 1,
Consider, Left hand derivative (L.H.D.) of the function
$L.H.D.=(gof)'{{x}^{-}}$
$\therefore L.H.D.=\dfrac{d}{dx}\sin \left( \sqrt{-x+1} \right)$
As we know the Formula: $\dfrac{d}{dx}\sin x=\cos x$
$\therefore L.H.D.=\cos \left( \sqrt{-x+1} \right)\dfrac{d}{dx}\left( \sqrt{-x+1} \right)$
As we know the Formula: $\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}$
$\therefore L.H.D.=\cos \left( \sqrt{-x+1} \right)\dfrac{-1}{2\sqrt{-x+1}}$

Simply put x=1 in above equation,
$\therefore {{\left[ L.H.D. \right]}_{x=1}}=\cos \left( \sqrt{-1+1} \right)\dfrac{1}{2\sqrt{-1+1}}$
$\therefore {{\left[ L.H.D. \right]}_{x=1}}=\cos \left( 0 \right)\dfrac{1}{2\sqrt{0}}$
$\therefore {{\left[ L.H.D. \right]}_{x=1}}=\infty$………………………………………. (3)
Consider, Right hand derivative (R.H.D.) of the function
$R.H.D.=(gof)'{{x}^{+}}$
$\therefore R.H.D.=\dfrac{d}{dx}\sin \left( \sqrt{x-1} \right)$
As we know the Formula: $\dfrac{d}{dx}\sin x=\cos x$,
$\therefore R.H.D.=\cos \left( \sqrt{x-1} \right)\dfrac{d}{dx}\left( \sqrt{x-1} \right)$
As we know the Formula: $\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{\sqrt{x}}$
$\therefore R.H.D.=\cos \left( \sqrt{x-1} \right)\dfrac{1}{2\sqrt{x-1}}$
Simply put x=1 in above equation,
$\therefore {{\left[ R.H.D \right]}_{x=1}}=\cos \left( \sqrt{1-1} \right)\dfrac{1}{2\sqrt{1-1}}$
$\therefore {{\left[ R.H.D \right]}_{x=1}}=\cos \left( \sqrt{0} \right)\dfrac{1}{2\sqrt{0}}$
$\therefore {{\left[ R.H.D \right]}_{x=1}}=\infty$…………………………………… (4)
From (3) and (4) we can write,
$L.H.D.=R.H.D.$
Therefore, the function $(gof)x$ is differentiable at 1.

Note: Don’t get confused if you are getting $\infty$ as a derivative at a particular point as it only means that the tangent is parallel to the y-axis. Students generally think that when they are getting derivative as $\infty$ , it might be wrong. But it is not the case. Derivative of a function can be 0, any real number or $\infty$ .