The value of `f'(a) = lim_(x->4)(2^x - 16)/(x - 4)`
Substituting x = 4 gives an indeterminate form `0/0` . This allows l'Hopital's rule to be used and the numerator and denominator are substituted by their derivatives.
`lim_(x->4) (ln 2*2^x)/1`
Substituting x = 4 gives `ln 2*16`
There can be many combinations of functions f(x) and values of a for which f'(a) = `ln 2*16` , one example is `f(x) = ln 2*x^2/2`
`f'(x) = ln 2*x`
`f'(16) = ln 2*16`
This gives one solution as `f(x) = ln 2*(x^2/2)` and `a = 16`