Question

Let f(x)=5x-4,when 0<x<1 and f(x)=4x^2-3x, when x>1,discuss the differentiability of f(x) at x=1?

Answer 1

`f(x)` is not differentiable at `x=1`

Explanation:

The definition of differentiable requires that the function be defined at the point in question. In this case `f(x)` is not defined at `x=1`

Note that if either definition were expanded to include `x=1`,
then the function would be differentiable,
since the slope of `f(x)` as `xrarr 1` for `0 < x <= 1` is equal to the slope when `xrarr1` for `x > 1`

For `f(x)=5x-4`
slope`= f'(x)=5`

For `f(x)=4x^2-3x`
slope`= f'(x)=8x-3` and as `xrarr1`, `f'(xrarr1)rarr5`

Answer 2

Please see below.

Explanation:

Regardless of which definition you use,

`f'(1) = lim_(hrarr0)(f(1+h)-f(1))/h`

or `f'(1) = lim_(xrarr1)(f(x)-f(1))/(x-1)`

As Alan P has pointed out, no limit exists because `f(1)` is not defined.

For further discussion of how to make a function similar to this one that DOES have a derivative at `1`, please see Alan P's answer.