# Math Help - solving a fonction problem

1. ## solving a fonction problem

i have the equation:
$
(f(x+h)-f(x))/h
$

$
f(x)= root(x+1)
$

so i get this

$(root(x+h+1)-root(x+1))/h$

when i check the answer i have this :

$1/(root(x+h+1)+root(x+1))$

i do not know how to go from what i have to the answer.

for some reason √ doesnt like it in math wrap.

2. Originally Posted by DAngel
i have the equation:
$
(f(x+h)-f(x))/h
$

$
f(x)= root(x+1)
$

so i get this

$(root(x+h+1)-root(x+1))/h$

when i check the answer i have this :

$1/(root(x+h+1)+root(x+1))$

i do not know how to go from what i have to the answer.

for some reason √ doesnt like it in math wrap.
$\displaystyle \frac{\sqrt{x + 1 + h} - \sqrt{x + 1}}{h}$

Try multiplying the top and bottom by $\displaystyle \sqrt{x + 1 + h} + \sqrt{x + h}$. This is called "rationalizing the numerator." (As opposed to rationalizing the denominator, which you are probably more familiar.)

-Dan

3. can you show me. im not sure what you mean. i dont understand how he goes from what i ahve to the answer. what i mean is the logic behind it.

4. Do what topsquark said! What do you get when you multiply
$\frac{\sqrt{x+1+h}- \sqrt{x+ 1}}{h}\frac{\sqrt{x+1+h}+ \sqrt{x+1}}{\sqrt{x+1+h}+ \sqrt{x+1}}$?

5. Originally Posted by DAngel
can you show me. im not sure what you mean. i dont understand how he goes from what i ahve to the answer. what i mean is the logic behind it.
I've never taken a course in Analysis, but my take on why we rationalize the numberator is simply that the trick works. Now that you've seen what to do with this problem you know that you can do it to others similar to it.

(Eventually you'll take the limit of your answer as h goes to 0. Then you'll see why this simplifies the problem rather than make it more difficult.)

-Dan

6. i must not see what you mean. i tried, all i end up is with the same equation since they just cancel each other out.
would you mind showing me the steps you used to make solve the problem?

7. Originally Posted by DAngel
i must not see what you mean. i tried, all i end up is with the same equation since they just cancel each other out.
would you mind showing me the steps you used to make solve the problem?
$\displaystyle \frac{\sqrt{x+1+h}- \sqrt{x+ 1}}{h}\cdot \frac{\sqrt{x+1+h}+ \sqrt{x+1}}{\sqrt{x+1+h}+ \sqrt{x+1}}$

$\displaystyle \frac{\left [ \sqrt{x+1+h}- \sqrt{x+ 1} ~ \right ] \left [ \sqrt{x+1+h}+ \sqrt{x+1} ~ \right ]}{h \left [ \sqrt{x+1+h}+ \sqrt{x+1} ~ \right ]}$

The numerator is in the form $\displaystyle (a - b)(a + b) = a^2 - b^2$

Does this help?

-Dan

8. Originally Posted by DAngel
i must not see what you mean. i tried, all i end up is with the same equation since they just cancel each other out.
would you mind showing me the steps you used to make solve the problem?
They only cancel each other out IF YOU CANCEL THEM! That might sound funny, but the point of us inserting the same factor on top and on bottom is to somehow combine/rearrange so that we get something ELSE to cancel...

"FOIL" out the top. You will have "h" remaining. Just h. This cancels with the h factor on bottom.

Then directly substituting h = 0 will give something meaningful, and the answer.

9. yes it does help. but for some reason i dont see how to do the rest.

if i have

$(a^2-b^2)/(h(a+b))$

im still stuck because i dont see how to go beyond that.

TheChaz, im unsure what you are trying to say.

10. Originally Posted by topsquark
$\displaystyle \frac{\sqrt{x+1+h}- \sqrt{x+ 1}}{h}\cdot \frac{\sqrt{x+1+h}+ \sqrt{x+1}}{\sqrt{x+1+h}+ \sqrt{x+1}}$

$\displaystyle \frac{\left [ \sqrt{x+1+h}- \sqrt{x+ 1} ~ \right ] \left [ \sqrt{x+1+h}+ \sqrt{x+1} ~ \right ]}{h \left [ \sqrt{x+1+h}+ \sqrt{x+1} ~ \right ]}$

$\displaystyle \frac{\left [ \sqrt{x+1+h}*\sqrt{x+1+h} - \sqrt{x + 1}*\sqrt{x + 1} ~ \right ] }{h \left [ \sqrt{x+1+h}+ \sqrt{x+1} ~ \right ]}
$

$\displaystyle \frac{(x + 1 + h) - (x + 1)}{h(\sqrt{x + 1 + h} + \sqrt{x + 1})}$

...
Now there is simply "h" on top and bottom. Cancel those.

$(root(a))^2 = a$
12. just FYI if you use (\sqrt{a})^2=a you will get $(\sqrt{a})^2=a$