1. ## Difference Quotient

Find the difference quotient for the given function.

f(x) = sqrt{x} + 1

2. Originally Posted by magentarita
Find the difference quotient for the given function.

f(x) = sqrt{x} + 1
The difference quotient is defined as $\frac{f\left(x+h\right)-f\left(x\right)}{h}$ (When you study calculus, taking the limit as h approaches zero of the difference quotient will yield the derivative).

Since $f\left(x\right)=\sqrt{x}+1$, we see that the difference quotient is $\frac{\sqrt{x+h}+1-\left(\sqrt{x}+1\right)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}$

We can try to manipulate this a bit.

Multiplying both numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$, we get $\frac{x+h-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}$ which yields $\color{red}\boxed{\frac{1}{\sqrt{x+h}+\sqrt{x}}}$

Does this make sense?

3. ## yes but.............

Originally Posted by Chris L T521
The difference quotient is defined as $\frac{f\left(x+h\right)-f\left(x\right)}{h}$ (When you study calculus, taking the limit as h approaches zero of the difference quotient will yield the derivative).

Since $f\left(x\right)=\sqrt{x}+1$, we see that the difference quotient is $\frac{\sqrt{x+h}+1-\left(\sqrt{x}+1\right)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}$

We can try to manipulate this a bit.

Multiplying both numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$, we get $\frac{x+h-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}$ which yields $\color{red}\boxed{\frac{1}{\sqrt{x+h}+\sqrt{x}}}$

Does this make sense?
Yes it makes sense but why multiply by the conjugate if the end result will yield two radicals in the denominator?

4. Originally Posted by magentarita
Yes it makes sense but why multiply by the conjugate if the end result will yield two radicals in the denominator?
Good Question!

At this stage, I believe it would be fine to leave it as $\frac{\sqrt{x+h}-\sqrt{x}}{h}$

But as you go further on in math, it will be a little clearer why you would get it into the final form I have it in.

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(As an aside)

When you're in calculus and you need to evaluate $\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$, it is best to give it in the final form I have, for you can then easily evaluate the limit [the reason why we manipulate the original limit by multiplying it by the conjugate of the numerator is to avoid dividing by zero].

This probably will make more sense when you get closer to studying calculus.

5. ## I see.........

Originally Posted by Chris L T521
Good Question!

At this stage, I believe it would be fine to leave it as $\frac{\sqrt{x+h}-\sqrt{x}}{h}$

But as you go further on in math, it will be a little clearer why you would get it into the final form I have it in.

-------------------------------------------------------------------------

(As an aside)

When you're in calculus and you need to evaluate $\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$, it is best to give it in the final form I have, for you can then easily evaluate the limit [the reason why we manipulate the original limit by multiplying it by the conjugate of the numerator is to avoid dividing by zero].

This probably will make more sense when you get closer to studying calculus.
Thanks for clearing things up for me.