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  • <img alt="\mathrm{ f(x)=\left\{\begin{array}{ll}|x-4| & \text { for } x \geq 1 \\ x^{3} / 2-x^{2}+3 x+1 / 2 & \text { for } x<1,\end{array}\right.}" src="https://entrancecorner.oncodecog

\mathrm{ f(x)=\left\{\begin{array}{ll}|x-4| & \text { for } x \geq 1 \\ x^{3} / 2-x^{2}+3 x+1 / 2 & \text { for } x<1,\end{array}\right.} then

Option: 1

f(x) is continuous at x=1 and at x=4


Option: 2

f(x) is differentiable at x=4


Option: 3

f(x) is continuous and differentiable at x=1


Option: 4

f(x) is only continuous at x=1.


Answers (1)

best_answer

Since \mathrm{g(x)=|x|}  is a continuous function and \mathrm{\lim _{x \rightarrow 1+} f(x)=3=\lim _{x \rightarrow 1-} f(x)} so \mathrm{f} is a continuous function. In particular \mathrm{f} is continuous at \mathrm{x=1} and \mathrm{x=4}. \mathrm{f} is clearly not differentiable at \mathrm{x= 4}. Since \mathrm{g(x)=|x|}  is not differentiable at \mathrm{x=0}.Now

\begin{aligned} f^{\prime}(1+) & =\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{|-3+h|-3}{h}=-1, \\ f^{\prime}(1-) & =\lim _{h \rightarrow 0^{-}} \frac{(1 / 2)(1+h)^{3}-(1+h)^{2}+3(1+h)+1 / 2-3}{h} \\ & =\lim _{h \rightarrow 0^{-}} \frac{(1 / 2)\left(h^{3}+3 h^{2}+3 h\right)-\left(h^{2}+2 h\right)+3 h}{h} \\ & =\frac{5}{2} . \end{aligned}
Hence f is not differentiable at  \mathrm{x=1}.

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Nehul

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