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 \[\infty \] .