Help with calculus please!

Question

If f(x) = 3a|4x – 4| – ax, where a is some constant not equal to zero, find f ′(1).

Question options:

1)	0
2)	DNE
3)	not enough information
4)	1

Answer 1

When you perform a function transformation, you need to get the function into a standard form to see the effects of the transformation. For instance, the transformations of y=sin(3x-3) cannot readily be seen but you can see them in y = sin[3(x-1)] (the period is 2π/3 and the horizontal shift is +1)

So let's factor out the 4 in this function:

f(x) = 3a|4x – 4| – ax becomes

f(x) = 3a|4(x – 1)| – ax

This tells us that the absolute value function (which normally has a sharp point at x = 0) is being shifted right by 1 unit (the sharp point is now at x = 1). It also has a vertical stretch factor of "3a" but that doesn't change the fact there is a sharp point at x = 1. And it also has a vertical shift by "ax" but, again, that does not change the fact that there is a sharp point at x = 1.

So, because the function has a sharp point at x = 1, f'(1) does not exist (DNE)