How to Prove That the Function is Not Differentiable ?
Here we are going to see how to prove that the function is not differentiable at the given point.
The function is differentiable from the left and right. As in the case of the existence of limits of a function at x_{0}, it follows that
exists if and only if both
exist and f' (x_{0}-) = f' (x_{0}+)
Hence
if and only if f' (x_{0}-) = f' (x_{0}+). If any one of the condition fails then f'(x) is not differentiable at x_{0.}
From the above statements, we come to know that if f' (x_{0}-) ≠ f' (x_{0}+), then we may decide that the function is not differentiable at x_{0}.
Question 1 :
Show that the following functions are not differentiable at the indicated value of x.
(i)
Solution :
f'(2^{-}) = lim _{x->2-} [(f(x) - f(2)) / (x - 2)]
= lim _{x->2-} [(-x + 2) - (-2 + 2)] / (x - 2)
= lim _{x->2-} -(x - 2) / (x - 2)
= lim _{x->2-} (-1)
= -1 -----(1)
f'(2^{+}) = lim _{x->2+} [(f(x) - f(2)) / (x - 2)]
= lim _{x->2+} [(2x - 4) - (4 - 4)] / (x - 2)
= lim _{x->2+} 2(x -2) / (x - 2)
= 2 -----(2)
f'(2^{-}) ≠ f'(2^{+})
Hence the given function is not differentiable at the point x = 2.
Solution :
f'(0^{-}) = lim _{x->0-} [(f(x) - f(0)) / (x - 0)]
= lim _{x->0-} [(3x) - 0] / (x - 0)
= lim _{x->0-} (3x/x)
= lim _{x->0-} 3
= 3 -----(1)
f'(0^{+}) = lim _{x->0+} [(f(x) - f(0)) / (x - 0)]
= lim _{x->0+} [(-4x) - 0] / (x - 0)
= lim _{x->0+} (-4x/x)
= lim _{x->0+} -4
= -4 -----(2)
f'(0^{-}) ≠ f'(0^{+})
Hence the given function is not differentiable at the point x = 0.
Question 2 :
The graph of f is shown below. State with reasons that x values (the numbers), at which f is not differentiable.
Solution :
If a function is continuous at a point, then it is not necessary that the function is differentiable at that point.
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 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}.
At x = 1 and x = 8, we get vertical tangent (or) sharp edge and sharp peak. So it is not differentiable at x = 1 and 8.
At x = 4, we hjave a hole. Hence it is not continuous at x = 4.
At x = 11, we have perpendicular tangent. So it is not differentiable at x = 11.
Question 3 :
If f(x) = |x + 100| + x^{2}, test whether f'(-100) exists.
Solution :
f'(-100^{-}) = lim _{x->-100-} [(f(x) - f(-100)) / (x - (-100))]
= lim _{x->-100-} [(-(x + 100)) + x^{2}) - 100^{2}] / (x + 100)
= lim _{x->-100-} [(-(x + 100)) + (x^{2} - 100^{2})] / (x + 100)
= lim _{x->-100-} [(-(x + 100)) + (x + 100) (x -100)] / (x + 100)
= lim _{x->-100-} (x + 100)) [-1 + (x -100)] / (x + 100)
= [-1 + (-100 -100)]
= -201 --------(1)
f'(-100^{+}) = lim _{x->-100+} [(f(x) - f(-100)) / (x - (-100))]
= lim _{x->-100-} [(x + 100)) + x^{2}) - 100^{2}] / (x + 100)
= lim _{x->-100-} [(x + 100)) + (x^{2} - 100^{2})] / (x + 100)
= lim _{x->-100-} [(x + 100)) + (x + 100) (x -100)] / (x + 100)
= lim _{x->-100-} (x + 100)) [1 + (x -100)] / (x + 100)
= lim _{x->-100-} (x - 99)
= -199 --------(2)
Hence f'(-100) does not exists.
Question 4 :
Examine the differentiability of functions in R by drawing the diagrams.
(i) | sin x |
Solution :
Graph of y = sin x
There is vertical tangent for nπ. Hence it is not differentiable at x = nπ, n ∈ z
(ii) | cos x|
Solution :
There is vertical tangent for (2n + 1)(π/2). Hence it is not differentiable at x = (2n + 1)(π/2), n ∈ z
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