Problems from Differentiability and its relation with continuity

Development:

To study the differentiability at $x=1$ we must do the side derivatives at $x=1$ and see if they exist and coincide.

a) $f(x)=x^3+x-2$

$$f^{\prime }(0^{-})=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}{\displaystyle \frac{f(1+\mathrm{\Delta }x)-f(1)}{\mathrm{\Delta }x}}=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}{\displaystyle \frac{(1+\mathrm{\Delta }x{)}^3+(1+\mathrm{\Delta }x)-2-1^3-1+2}{\mathrm{\Delta }x}}=$$ $$=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}\frac{1+3\mathrm{\Delta }x+3\mathrm{\Delta }{x}^2+\mathrm{\Delta }{x}^3+1+\mathrm{\Delta }x-2}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}{\displaystyle \frac{\mathrm{\Delta }{x}^3+3\mathrm{\Delta }{x}^2+4\mathrm{\Delta }x}{\mathrm{\Delta }x}}=$$ $$=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}(\mathrm{\Delta }{x}^2+3\mathrm{\Delta }x+4)=4$$

$$f^{\prime }(0^{+})=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}{\displaystyle \frac{f(1+\mathrm{\Delta }x)-f(1)}{\mathrm{\Delta }x}}=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}{\displaystyle \frac{(1+\mathrm{\Delta }x{)}^3+(1+\mathrm{\Delta }x)-2-1^3-1+2}{\mathrm{\Delta }x}}=$$ $$=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}\frac{1+3\mathrm{\Delta }x+3\mathrm{\Delta }{x}^2+\mathrm{\Delta }{x}^3+1+\mathrm{\Delta }x-2}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}{\displaystyle \frac{\mathrm{\Delta }{x}^3+3\mathrm{\Delta }{x}^2+4\mathrm{\Delta }x}{\mathrm{\Delta }x}}=$$ $$=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}(\mathrm{\Delta }{x}^2+3\mathrm{\Delta }x+4)=4$$

The two values coincide, both approaching from the right and from the left. Therefore, this function is differentiable at $x=1$.

As it is differentiable we can say that it is continuous.

b) $f(x)={\displaystyle \frac{1}{(x-1{)}^2}}$

$$f^{\prime }(0^{-})=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}{\displaystyle \frac{f(1+\mathrm{\Delta }x)-f(1)}{\mathrm{\Delta }x}}=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{(1+\mathrm{\Delta }x-1{)}^2}}-{\displaystyle \frac{1}{(1-1{)}^2}}}{\mathrm{\Delta }x}}=$$ $$=\underset{\mathrm{\Delta }x\to 0^{-}}{lim}({\displaystyle \frac{\mathrm{\Delta }x}{\mathrm{\Delta }{x}^2}}-{\displaystyle \frac{\mathrm{\Delta }x}{0}})=-\mathrm{\infty }$$

$$f^{\prime }(0^{+})=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}{\displaystyle \frac{f(1+\mathrm{\Delta }x)-f(1)}{\mathrm{\Delta }x}}=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}{\displaystyle \frac{{\displaystyle \frac{1}{(1+\mathrm{\Delta }x-1{)}^2}}-{\displaystyle \frac{1}{(1-1{)}^2}}}{\mathrm{\Delta }x}}=$$ $$=\underset{\mathrm{\Delta }x\to 0^{+}}{lim}({\displaystyle \frac{\mathrm{\Delta }x}{\mathrm{\Delta }{x}^2}}-{\displaystyle \frac{\mathrm{\Delta }x}{0}})=+\mathrm{\infty }$$

The two values do not coincide, thus the function is not differentiable at $x=1$.

We can't say anything about the continuity in $x=1$.

Solution:

The first function is differentiable at $x=1$.

The second one is not differentiable at $x=1$.

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