1. ## evaluate and simplify

hey guys, need help with a practise question... thank you!!

if
$
f(x) = \frac{2}{3x+4}
$

evaluate and simplify

$
\frac{(2+h)-f(2)}{h}
$

2. Originally Posted by jvignacio
hey guys, need help with a practise question... thank you!!

if
$
f(x) = \frac{2}{3x+4}
$

evaluate and simplify

$
\frac{(2+h)-f(2)}{h}
$
have you wriiten the question correctly?

first, $f(2) = \frac{2}{3(2) + 4} = \frac{1}{5}$

so, $
\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

$\lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}$

am i right?

3. Originally Posted by kalagota
have you wriiten the question correctly?

first, $f(2) = \frac{2}{3(2) + 4} = \frac{1}{5}$

so, $
\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

$\lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h}$

am i right?
yes sorry thats it! your right. thats the problem. how would i solve it like that?

4. it should be familiar to you that

$f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$

if not yet, all you have to do is to substitute $2+h$ on $x$ of $f(x)$ to get $f(2+h)$

then solve for that limit..

$f(2+h) = \frac{2}{3(2+h)+4} = \frac{2}{10 + 3h}$

so, $\lim_{h\rightarrow 0} \frac{\frac{2}{10 + 3h} - \frac{1}{5}}{h}$...

i hope, you can continue from here..

5. Originally Posted by kalagota
it should be familiar to you that

$f'(x_0) = \lim_{h\rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}$

if not yet, all you have to do is to substitute $2+h$ on $x$ of $f(x)$ to get $f(2+h)$

then solve for that limit..

$f(2+h) = \frac{2}{3(2+h)+4} = \frac{2}{10 + 3h}$

so, $\lim_{h\rightarrow 0} \frac{\frac{2}{10 + 3h} - \frac{1}{5}}{h}$...

i hope, you can continue from here..
okay i got upto..

$
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
$

am i right? not sure what to do after

6. Originally Posted by jvignacio
okay i got upto..

$
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
$

am i right? not sure what to do after

anyone?

7. Originally Posted by jvignacio
okay i got upto..

$
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
$

am i right? not sure what to do after
that should be negative..

$
\lim_{h\rightarrow 0} \frac{\frac{-3h}{50+15h}}{h} = \lim_{h\rightarrow 0} \frac{-3h}{50+15h} \cdot \frac{1}{h} = \lim_{h\rightarrow 0} \frac{-3}{50 + 15h} ...
$

8. Originally Posted by jvignacio
okay i got upto..

$
\lim_{h\rightarrow 0} \frac{\frac{3h}{50+15h}}{h}
$

am i right? not sure what to do after
$
-\ \frac{\frac{3h}{50+15h}}{h}=-\ \frac{3}{50+15h}
$

Then the limit of this as $h$ goes to $0$ is obtained just by putting $h=0$, so is: $-3/50$.

RonL