# How do I use one sided derivatives to show that the function f(x)=x^3 , x<=1 f(x)=3x , x>1 is not differentiable at x=1?

Jun 28, 2018

Actually, it seems to me that the function is differentiable

#### Explanation:

This is a piecewise function, i.e. a function defined with different rules in different intervals.

In this case, both rules ${x}^{3}$ and $3 x$ are polynomials, which means that they are continuous and differentiable infinite times.

But what does this mean? Well, a function is differentiable if the right and left derivatives exist, and are the same.

So, inside both domains $x \setminus \le 1$ and $x > 1$, whenever you try to compute left and right derivatives you will have

$\setminus {\lim}_{h \setminus \to {0}^{+}} \setminus \frac{f \left(x + h\right) - f \left(x\right)}{h} = \setminus {\lim}_{h \setminus \to {0}^{-}} \setminus \frac{f \left(x + h\right) - f \left(x\right)}{h}$

This means that the only point in which this equation may break is when you pass from one domain to the other, i.e. when you switch from one definition to the other.

So, if $x \setminus \ge 1$, the function behaves as ${x}^{3}$, which means that its derivative is $3 {x}^{2}$. When $x$ approaches $1$ (from the left), the derivative approaches $3$.

On the other hand, if $x > 1$, the function behaves as $3 x$, which means that its derivative is constantly $3$. When $x$ approaches $1$ (from the left), the derivative approaches $3$.

So, we've proven that the left and right derivative of $f \left(x\right)$ is the same for every $x$: if $x \ne 1$ we are inside a domain, and the functions are differentiable because they are polynomials.

If $x = 1$, direct calculation has shown that the left and right derivatives are still the same.