Difference Quotient

TimelordGurl · #1 $Old$ November 4th, 2009, 08:23 PM

I am new here but I have a question about this difference quotient. Hope you can help me!

f(x)=√5x
f(x)-f(5)/x-5

didn't know how to show it but f(x)-f(5) are all over x-5

So thank you if you can help me figure this out! I know the answer but I am not sure how to get to it....

HallsofIvy · #2 $Old$ November 5th, 2009, 03:23 AM

Quote:

Originally Posted by TimelordGurl $View Post$

I am new here but I have a question about this difference quotient. Hope you can help me!

f(x)=√5x
f(x)-f(5)/x-5

didn't know how to show it but f(x)-f(5) are all over x-5

So thank you if you can help me figure this out! I know the answer but I am not sure how to get to it....

Okay, so this is $\frac{f(x)- f(5)}{x-5}$ . Now replace "f(x)" and "f(5)" by this specific function.

But is $f(x)= (\sqrt{5})x$ or is $f(x)= \sqrt{5x}$ ?

If the first, $\frac{f(x)- f(5)}{x-5}= \frac{(\sqrt{5}x- \sqrt{5}}{x-5}$ $= \frac{\sqrt{5}(x-5)}{x-5}= \sqrt{5}$ .
(Which you should have expected- $y= (\sqrt{5})x$ is a linear function with slope $\sqrt{5}$ .)

If the second, $frac{f(x)- f(5)}{x-5}= \frac{\sqrt{5x}- \sqrt{5}}{x-5}$ which is harder to simplify. Try multiplying both numerator and denominator by $\sqrt{5x}+\sqrt{5}$ .