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y-x-4-6x-3x-2-x-lt-2-2x-2-14x-20-x-2-Is-the-function-y-x-differentiable-with-respect-to-x-at-x-2-




Question Number 13955 by ajfour last updated on 25/May/17
y(x)= { ((4+6x−3x^2        ;  x < 2)),((2x^2 −14x+20  ;  x ≥ 2)) :}  Is the function y(x)   differentiable with respect to x  at x=2 ?
$${y}\left({x}\right)=\begin{cases}{\mathrm{4}+\mathrm{6}{x}−\mathrm{3}{x}^{\mathrm{2}} \:\:\:\:\:\:\:;\:\:{x}\:<\:\mathrm{2}}\\{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{20}\:\:;\:\:{x}\:\geqslant\:\mathrm{2}}\end{cases} \\ $$$${Is}\:{the}\:{function}\:{y}\left({x}\right)\: \\ $$$${differentiable}\:{with}\:{respect}\:{to}\:{x} \\ $$$${at}\:{x}=\mathrm{2}\:? \\ $$
Commented by prakash jain last updated on 25/May/17
lim_(x→2^− )  y(x)=4+12−12=4  lim_(x→2^+ )  y(x)=8−28+20=0  y(x) is not continous at x=2.  Hence not differentiable.
$$\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:{y}\left({x}\right)=\mathrm{4}+\mathrm{12}−\mathrm{12}=\mathrm{4} \\ $$$$\underset{{x}\rightarrow\mathrm{2}^{+} } {\mathrm{lim}}\:{y}\left({x}\right)=\mathrm{8}−\mathrm{28}+\mathrm{20}=\mathrm{0} \\ $$$${y}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{continous}\:\mathrm{at}\:{x}=\mathrm{2}. \\ $$$$\mathrm{Hence}\:\mathrm{not}\:\mathrm{differentiable}. \\ $$
Commented by ajfour last updated on 25/May/17
very true, thanks Sir .
$${very}\:{true},\:{thanks}\:{Sir}\:. \\ $$

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