Question Number 214342 by issac last updated on 06/Dec/24
$$\mathrm{why} \\ $$$$\mathrm{differantiable}\:{f}\:\rightarrow\:{f}\:\mathrm{is}\:\mathrm{continious}\: \\ $$$$\mathrm{but}\:{f}\:\mathrm{is}\:\mathrm{continous}\:\nrightarrow\:\mathrm{differantiable}\:?? \\ $$
Commented by mr W last updated on 06/Dec/24
$${a}\:{smooth}\:{line}\:{is}\:{always}\:{continous}, \\ $$$${but}\:{a}\:{continous}\:{line}\:{is}\:{not}\:{always} \\ $$$${smooth}. \\ $$
Commented by mr W last updated on 06/Dec/24
Commented by mr W last updated on 06/Dec/24
$${all}\:{are}\:{continious},\:{but}\:{not}\:{all}\:{are} \\ $$$${smooth}\:\left({differentiable}\right). \\ $$
Answered by Frix last updated on 06/Dec/24
$$\mathrm{Just}\:\mathrm{think}\:\mathrm{of}\:\mathrm{an}\:\mathrm{example}: \\ $$$${f}\left({x}\right)=\mid{x}\mid\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{but}\:\mathrm{not}\:\mathrm{differentiable} \\ $$$$\mathrm{at}\:{x}=\mathrm{0} \\ $$$${g}\left({x}\right)=\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{but}\:\mathrm{not}\:\mathrm{differentiable} \\ $$$$\mathrm{at}\:{x}=\pm\mathrm{1} \\ $$