Use definition: f(x) = lim h→0 f(x + h) − f(x)/h to find f ' (x) and then find the tangent line to the graph of y=f(x) at x=a
f(x)=sqrt(x+1); a=8
I'm not sure what to do here.

Question

Use definition: f(x) = lim h→0 f(x + h) − f(x)/h    to find f ' (x) and then find the tangent line to the graph of y=f(x) at x=a
f(x)=sqrt(x+1); a=8
I'm not sure what to do here.

anonymous · Answer

plug ins   sqrt(x+1)   to your f(x) definition

 f(x) = lim h→0 f(x + h) − f(x)/h

--- or it could be easier to  plug in   a   from the start

plug ins   sqrt(8+1)   to your f(x) definition

 f(sqrt(8+1)) = lim h→0 f((sqrt(8+1)) + h) − f(sqrt(8+1))/h

amistre64 · Answer

thanks to the sqrt, conjugates are involved

wesdg1978 · Answer

OK, the answer is supposed to be 1/(2sqrt(x+1)) and y=1/6x+5/3 
I'm not seeing how to get there.

amistre64 · Answer

define the starting setup

apply conjugates

simplify out the h dividor

amistre64 · Answer

cancel out the h dividor?  .... simplify, cancel .... either way it gets taken care of

anonymous · Answer

I think i did a mistake in mines.

amistre64 · Answer

$$\lim~\frac{f(x+h)-f(x)}{h}$$
$$\lim~\frac{f(x+h)-f(x)}{h}\frac{f(x+h)+f(x)}{f(x+h)+f(x)}$$
$$\lim~\frac{f^2(x+h)-f^2(x)}{h(f(x+h)+f(x))}$$
with the given f(x) function, that undoes the sqrts and you can factor out an h above

amistre64 · Answer

as such, as h goes to 0$$\lim~\frac{1}{f(x+0)+f(x)}=\frac{1}{2f(x)}$$

wesdg1978 · Answer

OK, that makes sense, now I need to get the tangent line to the graph at x=a and a=8. How do I do that.

amistre64 · Answer

a line equation is from algebra:  the tangent line form will with be point slope, or slope intercept ... not that it really matter;  the results are the same regardless

y = f'(a)(x-a) + f(a)

amistre64 · Answer

once you know f',  the defines the slope for any given a

wesdg1978 · Answer

So, y=1/(2sqrt(x+1)) (x-8)+f(sqrt8+1) ?

amistre64 · Answer

close, define f'(a)  not f'(x),  and dont use f(sqrt(x+1))  thats nowhere defined .. other than being a typo  f(x) = sqrt(x+1)

y= 1/(2 sqrt(8+1)) (x-8) + sqrt(8+1) ?

wesdg1978 · Answer

So this is what I got from that: y=1/6x-8+3 or y=1/6x-5
The answer in the back of the book is y=1/6x+5/3

amistre64 · Answer

$$y=\frac{1}{2\sqrt9}(x-8)+\sqrt{9}$$
$$y=\frac{1}{2.3}(x-8)+3$$
$$y=\frac{1}{6}(x-8)+3$$
$$y=\frac{1}{6}x-\frac 86+3$$
$$y=\frac{1}{6}x-\frac 43+3$$
$$y=\frac{1}{6}x-\frac 43+\frac 93$$
$$y=\frac{1}{6}x+\frac 53$$

wesdg1978 · Answer

I just wasn't paying close enough attention, thanks for your help!

amistre64 · Answer

it happens ;)  good luck