Question

Draw the graph of the function \[y = |x - 1| + |x + 1|\] and discuss the continuity and differentiability of the function.

Answer 1

Hint: Here we simply draw the graph of the given function. A function is said to be continues if the curve has no missing points or breaking points in a given interval or domain. Function F(x) is said to be differentiable at a point then the differentiation of that function at that point exists. We know the definition of modulus of x.

Complete step by step answer:
We know have, \[|x| = \left\{ \begin{gathered}
  x{\text{ }}x \geqslant 0 \\
   - x{\text{ }}x < 0 \\
\end{gathered} \right.\] this is the standard definition.
We have \[|x - 1| = \left\{ \begin{gathered}
  x - 1{\text{ }}x > 1 \\
   - (x - 1){\text{ }}x < 1 \\
\end{gathered} \right.\] and \[|x + 1| = \left\{ \begin{gathered}
  x + 1{\text{ }}x > - 1 \\
   - (x + 1){\text{ }}x < - {\text{1}} \\
\end{gathered} \right.\]. ----- (1)
We have \[y = |x - 1| + |x + 1|\] ---- (2)
At \[x > 1\], the value of y will be
See the above definition (1) then we have,
\[ \Rightarrow y = x - 1 + x + 1 = 2x\] Is differentiable.
At \[x < - 1\],
By the definition (1) we will have,
\[ \Rightarrow y = - (x - 1) + ( - x - 1)\]
\[ \Rightarrow y = - x + 1 - x - 1\]
\[ \Rightarrow y = - 2x\] Is differentiable.
At \[x > 1\] in \[|x - 1|\] and \[x > - 1\]in \[|x + 1|\], then the value of y will be,
\[y = - (x + 1) + x + 1 = 2\] . Is differentiable.
We can see that y is differentiable,

Let’s draw the graph and check the continuity and differentiability.
At \[x = 0\] in equation (2) we have \[y = 2\]
Similarly, at \[x = 1\]\[ \Rightarrow y = 2\]
At \[x = - 1\]\[ \Rightarrow y = 2\]
At \[x = 2\]\[ \Rightarrow y = 4\]
At \[x = - 2\]\[ \Rightarrow y = 4\] and so on.

As we can see that the curve is continuous with no break points or missing points. That is \[y = |x - 1| + |x + 1|\] is continuous.
But while in differentiability, at point -1 and 1 the function \[y = |x - 1| + |x + 1|\] is not differentiable. Hence \[y = |x - 1| + |x + 1|\] is differentiable in \[R - \{ - 1,1\} \]. Or we can also say that it is differentiable at (-1, 1)

Note: We can directly plot the graph and by observation we can tell whether it is differentiable or continuous. The function is not differentiable at \[x = 1, - 1\] because the slopes at these points are different on the left and right hand side. In other wards Differentiability is defined as tangent to a curve.