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Show that f(x)=x^{3} is continuous as well as differentiable at x=3
CBSE | Class 12 | Continuity and Differentiability | Exercise 9C | Maths | RS Aggarwal
Show that f(x)=x^{3} is continuous as well as differentiable at x=3

Solution:

It is given that:
f(x)=x^{3}
If a function is differentiable at a point, it is necessary that it is continuous at that point. Left hand derivative at x=3
\begin{array}{l} \lim _{x \rightarrow 3^{-}} \frac{f(x)-f(3)}{x-3}=\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{(3-h)-3} \\ =\lim _{h \rightarrow 0} \frac{(3-h)^{3}-3^{3}}{(3-h)-3}=\lim _{h \rightarrow 0} \frac{(3-h)^{3}-27}{-h}=\lim _{h \rightarrow 0}-\frac{h\left\{(3-h)^{2}+3(3-h)+9\right\}}{h} \\ =\lim _{h \rightarrow 0}-\left\{(3-h)^{2}+3(3-h)+9\right\}=\lim _{h \rightarrow 0}-\left[-\left\{-(3-h)^{2}-3(3-h)-9\right\}\right] \end{array}
=l i m_{h \rightarrow 0}-\left\{-h^{2}+9 h-27\right\}=\lim _{h \rightarrow 0} h^{2}-9 h+27=0^{2}-9(0)+27=27
Right hand derivative at x=3
\begin{array}{l} \lim _{\mathrm{x} \rightarrow 3^{+}} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(3)}{\mathrm{x}-3}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(3+\mathrm{h})-\mathrm{f}(3)}{(3+\mathrm{h})-3} \\ =\lim _{\mathrm{h} \rightarrow 0} \frac{(3+\mathrm{h})^{3}-3^{3}}{(3+\mathrm{h})-3}=\lim _{\mathrm{h} \rightarrow 0} \frac{(3+\mathrm{h})^{3}-27}{\mathrm{~h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}\left[(3+\mathrm{h})^{2}+3(3+\mathrm{h})+9\right]}{\mathrm{h}} \\ =\lim _{\mathrm{h} \rightarrow 0}\left\{(3+\mathrm{h})^{2}+3(3+\mathrm{h})+9\right\}=\lim _{\mathrm{h} \rightarrow 0}(3+\mathrm{h})^{2}+3(3+\mathrm{h})+9 \\ =\lim _{\mathrm{h} \rightarrow 0}\left\{\mathrm{~h}^{2}+9 \mathrm{~h}+27\right\}=0^{2}+9(0)+27=27 \\ \mathrm{LHD}=\mathrm{RHD} \end{array}
So, f(x) is differentiable at x=3
\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3} x^{3}=3^{3}=27
Also, f(3)=27
Hence, f(x) is also continuous at x=3.

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