Question

Find f'(a) f(x)=square root of 1-2x using the equation:limit as x approaches 0 f(a+h)-f(a)/h

Answer 1

\[\Large f(\color{#DD4747 }{x})\quad=\quad \sqrt{1-2(\color{#DD4747 }{x})}\]

\[\Large f(\color{#DD4747 }{a})\quad=\quad \sqrt{1-2(\color{#DD4747 }{a})}\]\[\Large f(\color{#DD4747 }{a+h})\quad=\quad \sqrt{1-2(\color{#DD4747 }{a+h})}\]
Understand how function notation works?
We can now use these two pieces to plug them into our difference quotient.\[\Large \lim_{h\to0}\frac{f(a+h)-f(a)}{h}\]

Answer 2

I see some parts of it...did you send me a link? It looks like it but Im not sure?

Answer 3

A link? No.
Are the math equations not showing up?
Hmm you are using Internet Explorer I assume? :(

Answer 4

The math plugin doesn't work with IE :c

Answer 5

Oh ok will Mozilla fox work?

Answer 6

Wait, what is 'a' itself?

Answer 7

The derivative srry : (

Answer 8

Ok I see the picture now. I got to that point finally. What is the next thing I need to do?

Answer 9

So we plug in our pieces:
\[\Large \lim_{h\to0}\frac{\sqrt{1-2(\color{#DD4747 }{a+h})}-\sqrt{1-2(\color{#DD4747 }{a})}}{h}\]

Answer 10

Then we want to multiply the top and bottom by the `conjugate of the top`.

Answer 11

So we want to multiply the top and bottom by:\[\Large \left(\frac{\sqrt{1-2(\color{#DD4747 }{a+h})}+\sqrt{1-2(\color{#DD4747 }{a})}}{\sqrt{1-2(\color{#DD4747 }{a+h})}+\sqrt{1-2(\color{#DD4747 }{a})}}\right)\]

Answer 12

Ok I see that part there. So next we simplify or simplify it further?

Answer 13

Don't touch the denominator!
But yes we can simplify the numerator `quite a bit`.

Answer 14

Do you get 2h for the numerator?

Answer 15

Mmm yah that sounds right so far!

Answer 16

Well I probably got it the wrong way so please let me know...So I got -2h/h times square root 1-2a-2h + square root of 1-2a then I added up like terms and got -2h/h times square root of 2-4a-2h...then I just took out like terms or simplified it to -1/square root 1-2a...now I looked in the back of the book but I wanted to make sure I got it the right way..

Answer 17

Mmm I'm not quite sure what you're saying :(
The 2h sounded right, leaving you here,\[\Large \lim_{h\to0}\frac{2h}{h\left(\sqrt{1-2(\color{#DD4747 }{a+h})}+\sqrt{1-2(\color{#DD4747 }{a})}\right)}\]The h's cancel,\[\Large \lim_{h\to0}\frac{2}{\left(\sqrt{1-2(\color{#DD4747 }{a+h})}+\sqrt{1-2(\color{#DD4747 }{a})}\right)}\]

And now it's in a nice form where we can simply plug h=0 directly in.

Answer 18

Oh I missed a minus sign, woops\[\Large \lim_{h\to0}\frac{-2}{\left(\sqrt{1-2(\color{#DD4747 }{a+h})}+\sqrt{1-2(\color{#DD4747 }{a})}\right)}\]

Answer 19

It sounds like you got to the answer, or no?
I can't tell based on the way you described it :(

Answer 20

Yes but you explained it in a way that I'll copy the way you did it instead.haha

Answer 21

After you get your -2h on top, you cancel out an h from the top and bottom.
Then we plug in h=0,\[\Large \frac{-2}{\left(\sqrt{1-2(\color{#DD4747 }{a})}+\sqrt{1-2(\color{#DD4747 }{a})}\right)}\]And we can add the sqrt's,\[\Large \frac{-2}{2\sqrt{1-2a}}\]And then simplify a little further.

Answer 22

Ok I see it now thank u