Question

Use the definition limit as x approaches a(f(x)-f(a))/( x-a) to find the derivative of
f(x) = square root of x at x=3

Answer 1

\[\Large\rm f(x)=\sqrt x\]
\[\Large\rm f'(3)=\quad \lim_{x\to3}\frac{f(x)-f(3)}{x-3}\]

Answer 2

\[\Large\rm f(3)=\sqrt{3}\]

Answer 3

Understand how to plug the pieces into the formula?

Answer 4

Yes. I understand plugging in the 3 but I'm confused about the a

Answer 5

What do you do about the x?

Answer 6

a=3 for this problem.
So after we plug some stuff in,\[\Large\rm \lim_{x\to3}\frac{\sqrt x-\sqrt3}{x-3}\]We would ummmm, I got we need to do some fancy algebra steps.
Right now we're getting the indeterminate form 0/0.
So we'll start by multiplying the top and bottom by the `conjugate` of the numerator.

Answer 7

\[\Large\rm \lim_{x\to3}\frac{\sqrt x-\sqrt3}{x-3}\cdot\color{royalblue}{\left(\frac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}\right)}\]

Answer 8

\[\Large\rm \lim_{x\to3}\frac{(\sqrt x-\sqrt3)(\sqrt{x}+\sqrt{3})}{(x-3)(\sqrt{x}+\sqrt{3})}\]Leave the bottom alone, don't expand out the brackets.
But we DO want to expand out the top.
Remember what happens when you multiply conjugates?

Answer 9

you get x-3 at the top so it can cancel out with the bottom

Answer 10

That makes so much more sense. Thank you!!!

Answer 11

cool c: