Now: If f(x) is continuous at x = 0
$\underset{\mathit{x}\to 0^{+}}{\lim }\mathit{f}\left(\mathit{x}\right)=\underset{\mathit{x}\to 0^{-}}{\lim }\mathit{f}\left(\mathit{x}\right)=\mathit{f}\left(0\right)$
$\Rightarrow 1=\underset{\mathit{h}\to 0}{\lim }\mathit{f}\left(0-\mathit{h}\right)=1$
$\Rightarrow 1=\underset{\mathit{h}\to 0}{\lim }\left(-2\left(-\mathit{h}\right)+1\right)=1$
$\Rightarrow 1=1=1$
Which is true
$\Rightarrow \mathit{f}\left(\mathit{x}\right)$ is continuous at x = 0
Similarly for continuity at x = 1
$\underset{\mathit{h}\to 0}{\lim }\mathit{f}\left(\mathit{x}\right)=\underset{\mathit{x}\to 1^{+}}{\lim }\mathit{f}\left(\mathit{x}\right)=\mathit{f}\left(1\right)$
$\underset{\mathit{h}\to 0}{\lim }\mathit{f}\left(1-\mathit{h}\right)=\underset{\mathit{h}\to 0}{\lim }\mathit{f}\left(1+\mathit{h}\right)=\mathit{f}\left(1\right)$
$1=\lim {\mathrm{limits}}_{\mathit{h}\to 0}\left(2\left(1+\mathit{h}\right)-1\right)=1$
$\Rightarrow 1=1=1$
$\Rightarrow \mathit{f}\left(\mathit{x}\right)$ is continuous at x = 1
When x >0
$\underset{\mathit{x}\to 0^{+}}{\lim }{\mathit{f}}^{\prime }\left(\mathit{x}\right)=\underset{\mathit{x}\to 0}{\lim }\frac{\mathit{f}\left(0+\mathit{h}\right)-\mathit{f}\left(0\right)}{\mathit{h}}$
$=\underset{\mathit{h}\to 0}{\lim }\frac{1-1}{\mathit{h}}=0$
$\underset{\mathit{x}\to 0^{-}}{\lim }{\mathit{f}}^{\prime }\left(\mathit{x}\right)=\underset{\mathit{h}\to 0}{\lim }\frac{\mathit{f}\left(0-\mathit{h}\right)-\mathit{f}\left(0\right)}{-\mathit{h}}$
$=\underset{\mathit{h}\to 0}{\lim }\frac{-2\left(-\mathit{h}\right)+1-1}{-\mathit{h}}=-2$
$\therefore {\mathit{f}}^{\prime }\left(0^{+}\right)\ne {\mathit{f}}^{\prime }\left(0^{-}\right)$
∴ f(x) is not differentiable at x = 0
Now ${\mathit{f}}^{\prime }\left(1^{+}\right)=\underset{\mathit{h}\to 0}{\lim }\frac{\mathit{f}\left(1+\mathit{h}\right)-\mathit{f}\left(1\right)}{\mathit{h}}$
$=\underset{\mathit{h}\to 0}{\lim }\frac{2\left(1+\mathit{h}\right)-1-1}{\mathit{h}}$
$=\underset{\mathit{h}\to 0}{\lim }\frac{2\mathit{h}}{\mathit{h}}=2$
$\therefore {\mathit{f}}^{\prime }\left(1^{+}\right)\ne {\mathit{f}}^{\prime }\left(1^{-}\right)$
$\Rightarrow \mathit{f}\left(\mathit{x}\right)$ is not differentiable at x = 1
∴ f(x) is continuous at x = 0,1 and non-differentiable at x = 0,1