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

Oct 13, 2017

$f ' \left(x\right) = - \frac{1}{2}$
See below for details.

#### Explanation:

We will use the limit definition, $f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{9 - \frac{1}{2} \left(x + h\right) - \left(9 - \frac{1}{2} x\right)}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{9 - \frac{1}{2} x - \frac{1}{2} h - 9 + \frac{1}{2} x}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{- \frac{1}{2} h}{h}$

$f ' \left(x\right) = {\lim}_{h \to 0} - \frac{1}{2}$

$f ' \left(x\right) = - \frac{1}{2}$

Hopefully this helps!

Oct 13, 2017

This is the formula for finding the derivative using limits:

$\frac{d}{\mathrm{dx}} f \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x - h\right) - f \left(x\right)}{h}$

Plug in $\left(x - h\right)$ wherever you see an $x$ in your $f \left(x\right)$ 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:

$= {\lim}_{h \to 0} \frac{\left(9 - \frac{x - h}{2}\right) - \left(9 - \frac{x}{2}\right)}{h}$

$= {\lim}_{h \to 0} \frac{\left(\frac{18}{2} - \frac{x - h}{2}\right) - \left(\frac{18}{2} - \frac{x}{2}\right)}{h}$

$= {\lim}_{h \to 0} \frac{\left(\frac{18 - x + h}{2}\right) - \left(\frac{18 - x}{2}\right)}{h}$

$= {\lim}_{h \to 0} \frac{\frac{18 - x + h}{2} - \frac{18 + x}{2}}{h}$

$= {\lim}_{h \to 0} \frac{\frac{\cancel{18} \cancel{-} x + \cancel{h}}{2} - \frac{\cancel{18} + \cancel{x}}{2}}{\cancel{h}}$

$= {\lim}_{h \to 0} - \frac{1}{2}$

$= - \frac{1}{2}$