Differentiability and Continuity
Test the differentiability of the function f(x) = |x - 2| at x = 2.
We know that this function is continuous at x = 2.
Since the one sided derivatives f ′(2− ) and f ′(2+ ) are not equal, f ′ (2) does not exist. That is, f is not differentiable at x = 2. At all other points, the function is differentiable.
If x0 ≠ 2 is any other point then
The fact that f ′ (2) does not exist is reflected geometrically in the fact that the curve y = |x - 2| does not have a tangent line at (2, 0). Note that the curve has a sharp edge at (2, 0).
Examine the differentiability of f (x ) = x1/3 at x = 0.
Let f (x ) = x1/3. Clearly, there is no hole (or break) in the graph of this function and hence it is continuous at all points of its domain.
Let us check whether f ′(0) exists.
Therefore, the function is not differentiable at x = 0. From the Fig. 10.19, further we conclude that the tangent line is vertical at x = 0. So f is not differentiable at x = 0.
If a function is continuous at a point, then it is not necessary that the function is differentiable at that point.
What can you say about the differentiability of this function at other points?
Therefore f ′ (0) does not exist.
Here we observe that the graph of f has a jump at x = 0. That is x = 0 is a jump discontinuity.
The above illustrations and examples can be summarised to have the following conclusions.
A function f is not differentiable at a point x0 belonging to the domain of f if one of the following situations holds:
(i) f has a vertical tangent at x0.
(ii)The graph of f comes to a point at x0 (either a sharp edge ∨ or a sharp peak ∧ )
(iii) f is discontinuous at x0.
A function fails to be differentiable under the following situations :
If f is differentiable at a point x = x0, then f is continuous at x0.
Let f(x) be a differentiable function on an interval (a, b) containing the point x0. Then
This implies, f is continuous at x = x0.
The process of finding the derivative of a function using the conditions stated in the definition of derivatives is known as derivatives from first principle.
(1) Find the derivatives of the following functions using first principle.
(i) f (x) = 6 (ii) f(x) = - 4x + 7 (iii) f(x) = - x2 + 2
(2) Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?
(3) Determine whether the following function is differentiable at the indicated values.
(i) f(x) = x | x | at x = 0
(ii) f(x) = | x 2 - 1 | at x =1
(iii) f(x) = |x| + |x - 1| at x = 0, 1
(iv) f (x ) = sin | x | at x = 0
(4) Show that the following functions are not differentiable at the indicated value of x.
(5) The graph of f is shown below. State with reasons that x values (the numbers), at which f is not differentiable.
(6) If f(x) = |x + 100| + x2, test whether f ′(−100) exists.
(7) Examine the differentiability of functions in R by drawing the diagrams.
(i) | sin x | (ii) | cos x | .