Question

Question #ba1d3

Answer 1

e) `8x-3`

Explanation:

When you see (x+h) in the first principles formula, wherever you'd normally put `x` put `(x+h)`

`lim_(h->0)` `(4(x+h)^2-3(x+h)+5-(4x^2-3x+5))/h` substituting

`lim_(h->0)` `(4(x^2+h^2+2xh)-3x-3h+5-4x^2+3x-5)/h` expanding

`lim_(h->0)` `(4x^2+4h^2+8xh-3x-3h+5-4x^2+3x-5)/h`

quite a lot of these terms will cancel:

`lim_(h->0)` `(4h^2+8xh-3h)/h`

= `lim_(h->0)` `4h+8x-3`

now look at the limit - as h goes to zero, what happens?

`4(0)+8x-3 = 8x-3`