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Question Number 103998 by  M±th+et+s last updated on 18/Jul/20

$$iff\left(x\right)=x^{\frac{3}{2}}$$$$f^{\prime }\left(0\right)=0ornotexist$$

Commented by  M±th+et+s last updated on 19/Jul/20

Commented by  M±th+et+s last updated on 19/Jul/20

Commented by  M±th+et+s last updated on 19/Jul/20

After reviewing some books I found this

Commented by  M±th+et+s last updated on 19/Jul/20

Commented by  M±th+et+s last updated on 19/Jul/20

$$ipostedthisquestionfromthisexam$$$$$$$$andthanksforall$$

Answered by frc2crc last updated on 18/Jul/20

$$\frac{d}{dx}{x}^{3/2}=\frac{3}{2}{x}^{3/2-1}=\frac{3}{2}\sqrt{x}$$$$so{f}^{\prime }\left(0\right)=0$$

Commented by  M±th+et+s last updated on 18/Jul/20

$$thankyousir$$$$\underset{x\to 0^{+}}{limf}\left(x\right)=0but\underset{x\to 0^{-}}{limf}\left(x\right)isnotexist$$$$soithinkwecantfind{f}^{\prime }\left(0\right)$$

Commented by frc2crc last updated on 18/Jul/20

$$it{doesn}^{\prime }tmakeanysense$$$$\sqrt{x}is{}^{\prime }{only}^{\prime }forpositive$$$$numbers,whydoyouneed$$$$alimit?$$$$$$

Commented by  M±th+et+s last updated on 18/Jul/20

$$iaskedaboutthelimitbecause$$$$f^{\prime }\left(a\right)=\underset{x\to a}{lim}\frac{f\left(x\right)-f\left(a\right)}{x-a}$$$$f^{\prime }\left(0\right)=\underset{x\to 0}{lim}\frac{x\sqrt{x}-0}{x-0}=\underset{x\to 0}{lim}\sqrt{x}$$$$$$$$andi{wasn}^{,}tsureif\underset{x\to 0}{lim}\sqrt{x}=0$$$$$$$$andthankyouforyourexplaining$$$$$$

Answered by OlafThorendsen last updated on 18/Jul/20

$$\frac{f(0+h)-f\left(0\right)}{h}=\frac{h^{\frac{3}{2}}-0}{h}=\sqrt{h}$$$$\underset{h\to 0^{+}}{\lim }\frac{f(0+h)-f\left(0\right)}{h}=0$$$$\underset{h\to 0^{-}}{\lim }\frac{f(0+h)-f\left(0\right)}{h}\mathrm{does}\mathrm{not}\mathrm{exist}$$$$$$

Commented by  M±th+et+s last updated on 18/Jul/20

$$yessirthatwhatwasimean$$

Commented by bubugne last updated on 18/Jul/20

$$withk=-h$$$$h=i^{2}k(k⩾0)$$$$\underset{h\to 0^{-}}{\lim }\sqrt{h}=\underset{k\to 0^{+}}{\lim }i\sqrt{k}=i\times 0=0$$

Commented by prakash jain last updated on 18/Jul/20

$$\mathrm{If}\mathrm{the}\mathrm{domain}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{has}$$$$\mathrm{left}\mathrm{end}\mathrm{point}\mathrm{then}\mathrm{you}\mathrm{only}\mathrm{need}$$$$\mathrm{RHL}\mathrm{to}\mathrm{prove}\mathrm{continuity}.\mathrm{You}$$$$\mathrm{do}\mathrm{not}\mathrm{go}\mathrm{outside}\mathrm{of}\mathrm{domain}\mathrm{and}$$$$\mathrm{say}\mathrm{function}\mathrm{is}\mathrm{discontinuous}.$$$$f\left(x\right)=\sqrt{x}\mathrm{is}\mathrm{continuous}\mathrm{at}x=0.$$

Answered by Ar Brandon last updated on 18/Jul/20

$$\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{differentiable}\mathrm{at}\mathrm{a}\mathrm{point}{\mathrm{x}}_0\mathrm{iff}$$$$\underset{\mathrm{x}\to {\mathrm{x}}_0}{\lim }\frac{\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}\left({\mathrm{x}}_0\right)}{\mathrm{x}-{\mathrm{x}}_0}\mathrm{exists}.$$$$\underset{\mathrm{x}\to 0^{+}}{\lim }\frac{{\mathrm{x}}^{\frac{3}{2}}-0}{\mathrm{x}-0}=0,\underset{\mathrm{x}\to 0^{-}}{\lim }\frac{{\mathrm{x}}^{\frac{3}{2}}-0}{\mathrm{x}-0}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}.$$$$\mathrm{Therefore}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{not}\mathrm{differentiable}\mathrm{at}\mathrm{x}=0$$

Commented by prakash jain last updated on 18/Jul/20

$$$$$$\mathrm{If}\mathrm{the}\mathrm{domain}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{has}$$$$\mathrm{left}\mathrm{end}\mathrm{point}\mathrm{then}\mathrm{you}\mathrm{only}\mathrm{need}$$$$\mathrm{RHL}\mathrm{to}\mathrm{prove}\mathrm{continuity}.\mathrm{You}$$$$\mathrm{do}\mathrm{not}\mathrm{go}\mathrm{outside}\mathrm{of}\mathrm{domain}\mathrm{and}$$$$\mathrm{say}\mathrm{function}\mathrm{is}\mathrm{discontinuous}.$$$$f\left(x\right)=\sqrt{x}\mathrm{is}\mathrm{continuous}\mathrm{at}x=0.$$$$\mathrm{It}\mathrm{is}\mathrm{incorrect}\mathrm{to}\mathrm{take}\mathrm{LHL}\mathrm{for}$$$$\mathrm{function}\sqrt{x}$$

Commented by Ar Brandon last updated on 19/Jul/20

OK Sir. Thanks for the clarification.

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