Question

The function f is defined by y=f(x) where $x=2t-\left|t\right|,y=t^2+t\left|t\right|,tϵR.$ Draw the graph of f for the interval $-1\le x\le 1.$ Also discuss its continuity and differentiability at x=0.

• A. continuous for all
• B. differentiable for all
• C. dis -continuous at $x=0$
• D. not differentiable at $x=0$

Answer 1

Answer: (A), (B)

continuous for all
differentiable for all

Here $x=2t-\left|t\right|,y=t^2+t^2=2t^2$
Let $t\ge 0,thenx=2t-t=t,y=t^2+t^2=2t^2$ $\therefore y=2x^2,x\ge 0[\because t\ge 0\Rightarrow x\ge 0]$
and for $t\le 0,x=2t+t=3t,y=t^2-t^2=0$ $\therefore t=x/3andt\le 0\Rightarrow x\le 0$
Thus $\therefore t=\frac{x}{3}andt\le 0\Rightarrow x\le 0$
Hence the function is defined by $f\left(x\right)=2x^2,x\ge 0$ and $f\left(x\right)=y=0,x\le 0$
The function f is continuous at x=0,
Since $f(0+0)=\stackrel{Lt}{h\to 0}2{(0+h)}^2=0,$
$f(0-0)=0andf\left(0\right)=0$
f is also differentiable at x=0, for we have $R{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{2{(0+h)}^2-0}{h}=\underset{h\to 0}{lim}2h=0$
and $L{f}^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{0-0}{-h}=0.$
Thus the function f is continuous and differentiable for all real x.
The graph of the function is shown below.