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?

2 Answers
Dec 21, 2017

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

Explanation:

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

Note that if either definition were expanded to include x=1x=1,
then the function would be differentiable,
since the slope of f(x)f(x) as xrarr 1x1 for 0 < x <= 10<x1 is equal to the slope when xrarr1x1 for x > 1x>1

For f(x)=5x-4f(x)=5x4
slope = f'(x)=5

For f(x)=4x^2-3x
slope = f'(x)=8x-3 and as xrarr1, f'(xrarr1)rarr5
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Dec 21, 2017

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.