hey guys, need help with a practise question... thank you!!
if
$\displaystyle
f(x) = \frac{2}{3x+4}
$
evaluate and simplify
$\displaystyle
\frac{(2+h)-f(2)}{h}
$
have you wriiten the question correctly?
first, $\displaystyle f(2) = \frac{2}{3(2) + 4} = \frac{1}{5}$
so, $\displaystyle
\frac{(2+h)-f(2)}{h} = \frac{(2+h) + \frac{1}{5}}{h} = \frac{10 + 5h + 1}{5h} = \frac{11+5h}{5h}
$
but this looks familiar.. i was thinking that the problem was
$\displaystyle \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}$
am i right?
it should be familiar to you that
$\displaystyle f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$
if not yet, all you have to do is to substitute $\displaystyle 2+h$ on $\displaystyle x $ of $\displaystyle f(x)$ to get $\displaystyle f(2+h)$
then solve for that limit..
$\displaystyle f(2+h) = \frac{2}{3(2+h)+4} = \frac{2}{10 + 3h}$
so, $\displaystyle \lim_{h\rightarrow 0} \frac{\frac{2}{10 + 3h} - \frac{1}{5}}{h}$...
i hope, you can continue from here..