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

$$$f'(0^-)=\lim_{\Delta x \to 0^-}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^-}\dfrac{(1+\Delta x)^3+(1+\Delta x)-2-1^3-1+2}{\Delta x}=$$$ $$$=\lim_{\Delta x \to 0^-}\frac{1+3\Delta x+3\Delta x^2+\Delta x^3+1+\Delta x-2}{\Delta x}=\lim_{\Delta x \to 0^-}\dfrac{\Delta x^3+3\Delta x^2+4\Delta x}{\Delta x}=$$$ $$$=\lim_{\Delta x \to 0^-}(\Delta x^2+3\Delta x+4)=4$$$

$$$f'(0^+)=\lim_{\Delta x \to 0^+}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^+}\dfrac{(1+\Delta x)^3+(1+\Delta x)-2-1^3-1+2}{\Delta x}=$$$ $$$=\lim_{\Delta x \to 0^+}\frac{1+3\Delta x+3\Delta x^2+\Delta x^3+1+\Delta x-2}{\Delta x}=\lim_{\Delta x \to 0^+}\dfrac{\Delta x^3+3\Delta x^2+4\Delta x}{\Delta x}=$$$ $$$=\lim_{\Delta x \to 0^+}(\Delta x^2+3\Delta x+4)=4$$$

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) $$f(x)=\dfrac{1}{(x-1)^2}$$

$$$f'(0^-)=\lim_{\Delta x \to 0^-}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^-}\dfrac{\dfrac{1}{(1+\Delta x-1)^2}-\dfrac{1}{(1-1)^2}}{\Delta x}=$$$ $$$=\lim_{\Delta x \to 0^-}(\dfrac{\Delta x}{\Delta x^2}-\dfrac{\Delta x}{0})=-\infty$$$

$$$f'(0^+)=\lim_{\Delta x \to 0^+}\dfrac{f(1+\Delta x)-f(1)}{\Delta x}=\lim_{\Delta x \to 0^+}\dfrac{\dfrac{1}{(1+\Delta x-1)^2}-\dfrac{1}{(1-1)^2}}{\Delta x}=$$$ $$$=\lim_{\Delta x \to 0^+}(\dfrac{\Delta x}{\Delta x^2}-\dfrac{\Delta x}{0})=+\infty$$$

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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