Practice Questions on Differentiability and Continuity
Chapter: 11th Mathematics : UNIT 10 : Differential Calculus: Differentiability and Methods of Differentiation
If a function is continuous at a point, then it is not necessary that the function is differentiable at that point.
Differentiability and Continuity Illustration 10.3 Test the differentiability of the function f ( x ) = | x - 2| at x = 2. Solution 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 x 0 ≠ 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). Illustration 10.4 Examine the differentiability of f ( x ) = x 1/ 3 at x = 0. Solution Let f ( x ) = x 1/3 . 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? Illustration 10.5 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 x 0 belonging to the domain of f if one of the following situations holds: (i) f has a vertical tangent at x 0 . (ii)The graph of f comes to a point at x 0 (either a sharp edge ∨ or a sharp peak ∧ ) (iii) f is discontinuous at x 0 . A function fails to be differentiable under the following situations : If f is differentiable at a point x = x 0 , then f is continuous at x 0 . Let f ( x ) be a differentiable function on an interval ( a , b ) containing the point x 0 . Then This implies, f is continuous at x = x 0 . 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 ) = - 4 x + 7 (iii) f ( x ) = - x 2 + 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| + x 2 , test whether f ′(−100) exists. (7) Examine the differentiability of functions in R by drawing the diagrams. (i) | sin x | (ii) | cos x | .
We have seen in illustration 10.3 and 10.4, the function f ( x ) = | x -2| and f ( x ) = x 1/3 are respectively continuous at x = 2 and x = 0 but not differentiable there, whereas in Example 10.3 and Illustration 10.5, the functions
are respectively not continuous at any integer x = n and x = 0 respectively and not differentiable too. The above argument can be condensed and encapsuled to state: Discontinuity implies non-differentiability .
Theorem 10.1 (Differentiability implies continuity)
Proof
Derivatives from first principle
EXERCISE 10.1
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11th Mathematics : UNIT 10 : Differential Calculus: Differentiability and Methods of Differentiation : Differentiability and Continuity | Solved Example Problems, Exercise | Mathematics
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