Find the Derivatives From the Left and Right at the Given Point :
Here we are going to see how to find the derivatives from the left and right at the given point.
For a function y = f(x) defined in an open interval (a, b) containing the point x_{0}, the left hand and right hand derivatives of f at x = h are respectively denoted by f'(h^{-}) and f'(h^{+})
f'(h^{-}) = lim _{h-> 0}^{-}[f(x + h) - f(x)] / h
f'(h^{+}) = lim _{h-> 0}^{+}[f(x + h) - f(x)] / h
provided the limits exist.
Question 1 :
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?
(i) f(x) = |x - 1|
Solution :
If the function is differentiable, then
f'(1^{-}) = f'(1^{+})
f'(1^{-}) = lim_{x->1-} [f(x) - f(1)] / (x - 1)
= lim_{x->1- }[-(x - 1) - 0]/(x - 1)
= -1
f'(1^{+}) = lim_{x->1+ } [f(x) - f(1)] / (x - 1)
= lim_{x->1+ }[(x - 1) - 0]/(x - 1)
= 1
Hence the given function is not differentiable at x = 1.
(ii) f(x) = √(1 - x^{2})
Solution :
If the function is differentiable, then
f'(1^{-}) = f'(1^{+})
f'(1^{-}) = [f(x) - f(1)] / (x - 1)
= lim_{x->1- }[√(1 - x^{2}) - 0]/(x - 1)
= lim_{x->1- }[√(1 - x^{2}) - 0]/(1 - x)
= lim_{x->1- }-√(1 + x) / √(1 - x)
= -√2 / 0
= -∞
Hence the given function is not differentiable at x = 1.
Solution :
If the function is differentiable, then
f'(1^{-}) = f'(1^{+})
f'(1^{-}) = lim_{x->1-} [f(x) - f(1)] / (x - 1)
= lim_{x->1- }(x - 1)/(x - 1)
= 1
f'(1^{+}) = lim_{x->1+ } [f(x) - f(1)] / (x - 1)
= lim_{x->1+ }(x^{2} - 1)/(x - 1)
= lim_{x->1+ }(x + 1)(x - 1)/(x - 1)
= lim_{x->1+ }(x + 1)
= 2
f'(1^{-}) = 1 and f'(1^{+}) = 2, so the given function is not differentiable at x = 1.
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