Question
Show that the function is continuous everywhere but not differentiable at and .
Solution — Step by Step
For : both and , so .
For : but , so .
For : both positive, so .
. . . All three match, so is continuous at . Similarly at : from the left, from the right.
Left-hand derivative: on , so LHD at is .
Right-hand derivative: on , so RHD at is .
Since LHD RHD, is not differentiable at .
LHD on : , so LHD at is .
RHD on : , so RHD at is .
LHD RHD, so is not differentiable at .
Final Answer: is continuous everywhere but not differentiable at and .
Why This Works
A function with a “corner” or “kink” can be continuous (no jump) yet not differentiable (the left and right slopes disagree). The absolute-value function always has a corner at , so any sum of such terms has potential non-differentiability at each .
The piecewise rewriting is the workhorse: it converts a single ugly expression into three smooth pieces, on each of which the derivative is trivial. Then we just check the joints.
Alternative Method
Use the fact that is non-differentiable only at . The sum inherits non-differentiability from each summand at the corresponding point. No piecewise expansion needed if we trust this property.
Concluding “not continuous because not differentiable” reverses the actual implication. Differentiability implies continuity, not the other way around. A function can be continuous everywhere yet differentiable nowhere (Weierstrass function).
For sums/products/compositions of |·| functions, list all the “kink points” first. The function is continuous at every kink (assuming finite limits) but possibly non-differentiable there. Drill this for JEE.