Question

Draw the graph of the function f(x) = $x-|x-x^2|-1\le x\le 1$ & discuss the continuity or discontinuity of f in the interval $-1\le x\le 1$

• A. f is continuous
• B. f is discontinuous
• C. f is only piecewise continuous
• D. f is not defined in this interval

Answer 1

Answer: (A)

f is continuous

$f\left(x\right)=x-|x-x^2|$

$x-x^2>0$

$x(1-x)>0\Rightarrow x(x-1)<0$

$xϵ(0,1)$

$f\left(x\right)=\{\begin{array}{c}x+(x-x^2);-1\le x<0\\ 0;x=0\\ x+(x-x^2);0<x\le 1\end{array}$

$f\left(x\right)=\{\begin{array}{c}2x-x^2;-1\le x<0\\ 0;x=0\\ x^2;0<x\le 1\end{array}$

at $x=0$

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}(2x-x^2)=0$

${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\left(x^2\right)=0$

LHL$=$RHL $=f\left(0\right)$

$\Rightarrow$It is continuous