# 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.