Differentiability Examples

DIFFERENTIABILITY EXAMPLES

Example 1 : If f(x + y) = f(x) + f(y) - 2xy - 1 for all x and y. If f'(0) exists and f'(0) = -sin\(\alpha\), then find f{f'(0)}.

Solution : f'(x) = \(\displaystyle{\lim_{h \to 0}}\) \(f(x+h) - f(x)\over h\)

= \(\displaystyle{\lim_{h \to 0}}\) \({f(x)+f(h)-2xh-1} - f(x)\over h\)

= \(\displaystyle{\lim_{h \to 0}}\) -2x + \(\displaystyle{\lim_{h \to 0}}\) \(f(h)-1\over h\) = -2x + \(\displaystyle{\lim_{h \to 0}}\) \(f(h) - f(0)\over h\)

[Putting x = 0 = y in the given relation we find f(0) = f(0) + f(0) + 0 - 1 \(\implies\) f(0) = 1]

\(\therefore\)   f'(x) = -2x + f'(0) = -2x - sin\(\alpha\)

\(\implies\)   f(x) = -\(x^2\) - (sin\(\alpha\)).x + c

      f(0) = - 0 - 0 + c \(\implies\) c = 1

\(\therefore\)   f(x) = -\(x^2\) - (sin\(\alpha\)).x + 1

so, f{f'(0)} = f(-sin\(\alpha\)) = -\(sin^2\alpha\) + \(sin^2\alpha\) + 1 = 1



Example 2 : Discuss the continuity and differentiability of the function y = f(x) defined parametrically; x = 2t - |t-1| and y = 2\(t^2\) + t|t|.

Solution : Here x = 2t - |t-1| and y = 2\(t^2\) + t|t|.

Now when t < 0;

x = 2t - {-(t-1)} = 3t - 1 and y = 2\(t^2\) - \(t^2\) = \(t^2\) \(\implies\) = y = \({1\over 9}{(x+1)}^2\)

= when 0 \(\le\) t < 1

x = 2t - {-(t-1)} = 3t - 1 and y = 2\(t^2\) - \(t^2\) = 3\(t^2\) \(\implies\) = y = \({1\over 3}{(x+1)}^2\)

when t \(\ge\) 1;

x = 2t - (t-1) = t + 1 and y = 2\(t^2\) + \(t^2\) = 3\(t^2\) \(\implies\) = y = 3\({(x+1)}^2\)

Thus, y = f(x) = \({1\over 9}{(x+1)}^2\), x < -1

y = f(x) = \({1\over 3}{(x+1)}^2\), -1\(\le\)x < 2

y = f(x) = 3\({(x+1)}^2\), x\(\ge\) 2

We have to check differentiability at x = -1 and 2.

Differentiabilty at x = -1; LHD = f'(\(-1)^-\) = \(\displaystyle{\lim_{h \to 0}}\)\(f(-1-h) - f(-1)\over -h\) = \(\displaystyle{\lim_{h \to 0}}\) \({1\over 9}(-1-h+1)^2 - 0\over -h\) = 0

RHD = f'(\(-1)^+\) = \(\displaystyle{\lim_{h \to 0}}\)\(f(-1+h) - f(-1)\over h\) = \(\displaystyle{\lim_{h \to 0}}\) \({1\over 3}(-1+h+1)^2 - 0\over h\) = 0

Hence f(x) is differentiable at x = -1

\(\implies\)     continuous at x = -1.

To check differentiability at x = 2;

LHD = f'(\(2)^-\) = \(\displaystyle{\lim_{h \to 0}}\)\({1\over 3}(2-h+1)^2 - 3\over -h\) = 2

RHD = f'(\(2)^+\) = \(\displaystyle{\lim_{h \to 0}}\)\(3(2+h-1)^2 - 3\over h\) = 6

Hence f(x) is not differentiable at x = 2.

But continuous at x = 2, because LHD and RHD both are finite.

f(x) is continuous for all x and differentiable for all x, except x = 2.

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