Question

How do you find the derivative using limits of `f(x)=9-1/2x`?

Answer 1

`f'(x) = -1/2`
See below for details.

Explanation:

We will use the limit definition, `f'(x) = lim_(h->0) (f(x + h) - f(x))/h`.

`f'(x) = lim_(h->0) (9 - 1/2(x + h) - (9 - 1/2x))/h`

`f'(x) = lim_(h->0) (9- 1/2x - 1/2h - 9 + 1/2x)/h`

`f'(x) = lim_(h->0) (-1/2h)/h`

`f'(x) = lim_(h->0) -1/2`

`f'(x) = -1/2`

Hopefully this helps!

Answer 2

This is the formula for finding the derivative using limits:

`d/dxf(x) = lim_(h->0)(f(x-h)-f(x))/h`

Plug in `(x-h)` wherever you see an `x` in your `f(x)` for the first part of your numerator and then just the function itself for the second part of the numerator.

Then, its just basic arithmetic after that:

`=lim_(h->0)((9-(x-h)/2)-(9-x/2))/h`

`=lim_(h->0)((18/2-(x-h)/2)-(18/2-x/2))/h`

`=lim_(h->0)(((18-x+h)/2)-((18-x)/2))/h`

`=lim_(h->0)((18-x+h)/2-(18+x)/2)/h`

`=lim_(h->0)((cancel18cancel-x+cancelh)/2-(cancel18+cancelx)/2)/cancelh`

`=lim_(h->0)-1/2`

`=-1/2`