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.
