Question

If the function is given as $f\left( x \right)=\left\{ \begin{matrix}
-x-\dfrac{\pi }{2}, & \forall x\le -\dfrac{\pi }{2} \\
-\cos x, & \forall -\dfrac{\pi }{2} < x\le 0 \\
x-1 & \forall 0 < x\le 1 \\
\ln x, & \forall x > 1 \\
\end{matrix} \right.$, then
A. $f\left( x \right)$ is continuous at $x=-\dfrac{\pi }{2}$
B. $f\left( x \right)$ is not differentiable at $x=0$
C. $f\left( x \right)$ is differentiable at $x=1$
D. $f\left( x \right)$ is differentiable at $x=-\dfrac{3}{2}$

Answer 1

Hint: We have to find the continuity and the differentiability of the given piecewise function at certain points. We use the points and use their close values to find the different functions that will be available to operate. We take the final conclusion depending on the equality of the theorems $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=f\left( a \right)$ for continuity and $\displaystyle \lim_{x \to {{a}^{+}}}{{f}^{'}}\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}{{f}^{'}}\left( x \right)$ for differentiability.

Complete step by step solution:
We have to show the continuity and the differentiability of the given function at certain points.
For the function if the condition $\displaystyle \lim_{x \to {{a}^{+}}}f\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}f\left( x \right)=f\left( a \right)$ satisfies then it will continuous.
At $x=-\dfrac{\pi }{2}$, we take
\[\displaystyle \lim_{x \to {{\left( -\dfrac{\pi }{2} \right)}^{-}}}f\left( x \right)=f\left( -\dfrac{\pi }{2} \right)={{\left[ -x-\dfrac{\pi }{2} \right]}_{x=\left( -\dfrac{\pi }{2} \right)}}=0\]
$\displaystyle \lim_{x \to {{\left( -\dfrac{\pi }{2} \right)}^{+}}}f\left( x \right)={{\left[ -\cos x \right]}_{x=\left( -\dfrac{\pi }{2} \right)}}=0$
Therefore, $f\left( x \right)$ is continuous at $x=-\dfrac{\pi }{2}$ as \[\displaystyle \lim_{x \to {{\left( -\dfrac{\pi }{2} \right)}^{-}}}f\left( x \right)=f\left( -\dfrac{\pi }{2} \right)=\displaystyle \lim_{x \to {{\left( -\dfrac{\pi }{2} \right)}^{+}}}f\left( x \right)=0\].
For the function if the condition $\displaystyle \lim_{x \to {{a}^{+}}}{{f}^{'}}\left( x \right)=\displaystyle \lim_{x \to {{a}^{-}}}{{f}^{'}}\left( x \right)$ satisfies then it will differentiable.
Here ${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ f\left( x \right) \right]$.
We check differentiability at three different points of $x=-\dfrac{3}{2},0,1$.
At $x=-\dfrac{3}{2}$, we take
\[\displaystyle \lim_{x \to {{\left( -\dfrac{3 }{2} \right)}^{-}}}{{f}^{'}}\left( x \right)=\displaystyle \lim_{x \to {{\left( -\dfrac{3 }{2} \right)}^{+}}}{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( -x-\dfrac{3 }{2} \right) \right]}_{x=\left( -\dfrac{3 }{2} \right)}}=-1\]
Therefore, $f\left( x \right)$ is differentiable at $x=-\dfrac{3}{2}$ as \[\displaystyle \lim_{x \to {{\left( -\dfrac{3 }{2} \right)}^{-}}}{{f}^{'}}\left( x \right)=\displaystyle \lim_{x \to {{\left( -\dfrac{3 }{2} \right)}^{+}}}{{f}^{'}}\left( x \right)=-1\].
At $x=0$, we take
\[\displaystyle \lim_{x \to {{0}^{-}}}{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( -\cos x \right) \right]}_{x=0}}={{\left[ \sin x \right]}_{x=0}}=0\]
\[\displaystyle \lim_{x \to {{0}^{+}}}{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( x-1 \right) \right]}_{x=0}}=1\]
Therefore, $f\left( x \right)$ is not differentiable at $x=0$ as \[\displaystyle \lim_{x \to {{0}^{-}}}{{f}^{'}}\left( x \right)\ne \displaystyle \lim_{x \to {{0}^{+}}}{{f}^{'}}\left( x \right)\].
At $x=1$, we take
\[\displaystyle \lim_{x \to {{1}^{-}}}{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( x-1 \right) \right]}_{x=1}}=1\]
\[\displaystyle \lim_{x \to {{1}^{+}}}{{f}^{'}}\left( x \right)={{\left[ \dfrac{d}{dx}\left( \ln x \right) \right]}_{x=1}}={{\left[ \dfrac{1}{x} \right]}_{x=1}}=1\]
Therefore, $f\left( x \right)$ is differentiable at $x=1$ as \[\displaystyle \lim_{x \to {{1}^{-}}}{{f}^{'}}\left( x \right)=\displaystyle \lim_{x \to {{1}^{+}}}{{f}^{'}}\left( x \right)=1\].

Note: This type of differentiability checking is called differentiability of piecewise function. A piecewise function is differentiable at a point if both of the pieces have derivatives at that point, and the derivatives are equal at that point.