Question

Find f'(x) then find the tangent line to the graph of y=f(x) at x=a. Please see below for full formula

Answer 1

Use \[\lim_{h \rightarrow 0}\frac{ f(x+h)-f(x) }{ h }\]
to find f'(x) and then find the tangent line to the graph of y=f(x) at x=a.

Answer 2

i think the answer would depend largely on \(f(x)\) and also on \(a\)

Answer 3

Oh, yea and \[f(x)=\sqrt{x+1}\]
a=8

Answer 4

write
\[\lim_{h\to 0}\frac{\sqrt{8+1+h}-\sqrt{8+1}}{h}\] as a first step
then
\[\lim_{h\to 0}\frac{\sqrt{9+h}-3}{h}\] as a second step

Answer 5

the h just goes into the square root like that? and i thought you have to keep the x in the equation. confused.

Answer 6

then multiply top and bottom by the conjugate of \(\sqrt{9+h}-3\) which is \(\sqrt{9+h}+3\)

Answer 7

you are told what \(x\) is, it is \(8\) so i replaced \(x\) by \(8\)

Answer 8

you can do it with \(x\) if you like, and then plug in the 8 later
you will get the same answer

Answer 9

but first i need to find the equation of the tangent line, so can you keep the x in there. and im more confused about the h.

Answer 10

ok what do you need to find an equation of a line?

Answer 11

you need the formula. lol

Answer 12

actually you need two pieces of information
a point and a slope
we can find the point easy enough, since you are told \(a=8\) and \(f(x)=\sqrt{x+1}\) making \(f(8)=\sqrt{8+1}=3\) and the point is therefore \((8,3)\)

Answer 13

the second piece of information you need is the slope
that is what we are finding now

Answer 14

ok, so like will this be the starting point?