Difference quotient of f(x) = √x

[kuahji]

The question is "find and simplify the difference quotient."

Given function

f(x)=sq root of x

So what I did is insert (x+h) under the radical & got

sq root of (x+h), then I subtracted the sq root of x (original function)

My answer was sq root [(x+h) - sq root (x)] / h

The answer in the back of the book is 1 / [sq root (x+h) + sq root (x)]

I'm not understanding where I'm going wrong, the other problems didn't give me a problem, just this one.

 

[skeeter]

multiply numerator and denominator by [sqrt(x+h) + sqrt(x)] ...

[sqrt(x+h) - sqrt(x)]/h * [sqrt(x+h) + sqrt(x)]/[sqrt(x+h) + sqrt(x)] =

[(x+h) - x]/[h[sqrt(x+h) + sqrt(x)]] =

h/[h[sqrt(x+h) + sqrt(x)]] =

1/[sqrt(x+h) + sqrt(x)]

 

[kuahji]

Ok, thanks!

So is there a reason why its listed in that form instead the other? I think that is what I'm not understanding.

 

[skeeter]

you'll find out why when you have to take the limit of the difference quotient as h -> 0 ... it's a calculus concept.

 

[HallsofIvy]

The question is "find and simplify the difference quotient."

Given function

f(x)=sq root of x

So what I did is insert (x+h) under the radical & got

sq root of (x+h), then I subtracted the sq root of x (original function)

My answer was sq root [(x+h) - sq root (x)] / h

Not sure if this is a typo of a major misunderstanding. What you said you did is correct but you should have [sq root(x+h)- sq root(x)]/h. Do you see the difference? That is:
$$\frac{\sqrt{x+h}- \sqrt{x}}{h}$$
where yours is
$$\frac{sqrt{x+h- sqrt{x}}}{h}$$

The answer in the back of the book is 1 / [sq root (x+h) + sq root (x)]

I'm not understanding where I'm going wrong, the other problems didn't give me a problem, just this one.

As was said before, rationalize the numerator: multiply numerator and denominator by sq root(x+h)+ sqrt(x).
$$\frac{\sqrt{x+h}-sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+y}+\sqrt{x}}$$