Question

Small Question (formula related). Finding equation of tangent line at the given point.

Answer 1

Here are the formulas: \[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\] (#1)

Answer 2

\[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h } \]
(#2)

Answer 3

Two example problems:
(A) \[y=x^3-3x+1; Point (3,19) \]
(B)
\[y=\sqrt{x} ; Point (25,5)\]

Answer 4

For problem (A), it uses formula (#2). For problem (B), it uses formula (#1). How do I know what formulas to use, so I won't use formula #1 on example A and formula #2 on example B?

Answer 5

you can use either 1 or 2

Answer 6

yea..

Answer 7

Ah I see, so either can work.

Answer 8

i used to get familiar with equation 2 for all the problems when i started calc

Answer 9

equation 2 is derived by equation 1. let h=x-a

Answer 10

its just a slightly more neater way

Answer 11

I'm sure freckles will give a more clear view

Answer 12

example:
\[\lim_{x \rightarrow 3} \frac{f(x)-f(3)}{x-3} \\ =\lim_{x \rightarrow 3} \frac{(x^3-3x+1)-(3^3-3(3)+1)}{x-3} \\ =\lim_{x \rightarrow 3} \frac{(x^3-3^3)+(-3x+3(3))+(1-1)}{x-3 } \\ =\lim_{x \rightarrow 3} \frac{(x-3)(x^2+3x+3^2)-3(x-3)}{x-3} \\ \text{ divide \top and bottom by } x-3\]
\[\lim_{x \rightarrow 3} \frac{(x^2+3x+3^2)-3}{1} =\lim_{x \rightarrow 3}(x^2+3x+3^2-3) \\ \text{ plug in 3} \\ 3^2+3(3)+3^2-3=3(3^2)-3=27-3=24\]
I hope it is obvious I used the difference of cubes formula above

Answer 13

oops and that example was for f(x)=x^3-3x+1 at x=3 :

Answer 14

\[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\]
let some variable \[h=x-a\]
Substitute 'h' into the first equation (note x=h+a)
\[\lim_{h-a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }\]
\[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\]

Answer 15

both are inter-related

Answer 16

oops

Answer 17

i think i made a mistake

Answer 18

second, I need "checking" on this problem.

Answer 19

should be h+a==>a

Answer 20

\[\lim_{h+a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }\]

Answer 21

\[\lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \\ \text{ multiply top and bottom by top's conjugate } \\ \lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \cdot \frac{\sqrt{25+h}+\sqrt{25}}{\sqrt{25+h}+\sqrt{25}} \\ \lim_{h \rightarrow 0} \frac{(25+h)-(25)}{h(\sqrt{25+h}+\sqrt{25})} \\ \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{25+h}+\sqrt{25})} \\ \frac{h}{h}=1 \text{ so } \\ \lim_{h \rightarrow 0} \frac{1}{\sqrt{25+h}+\sqrt{25}}=\frac{1}{\sqrt{25}+\sqrt{25}} =\frac{1}{2 \sqrt{25}} =\frac{1}{2(5)}=\frac{1}{10}\]
and this is using the other route for the other function you had there at x=25

Answer 22

though yeah to prove they are the same definition and you can use a substitution like @chris00 is doing above

Answer 23

If I use formula #2 for example B.
\[\lim_{h \rightarrow 0}\frac{ f(25+h)-f(25) }{ h }\]
=\[\lim_{h \rightarrow 0}\frac{ \sqrt{25+h}-\sqrt{25} }{ h } \]
=\[\frac{ \sqrt{25+h}-5 }{ h }\times \frac{ \sqrt{25+h}+5 }{\sqrt{25+h} +5 }\]
=\[\lim_{h \rightarrow 0}\frac{ 1 }{ \sqrt{25+h}+5 }\]

Answer 24

yes!!!!
replace h with 0 now

Answer 25

plug in 0 to get 1/10

Answer 26

yep yep

Answer 27

but for some reason, it counts that as wrong answer

Answer 28

what is the question?

Answer 29

exactly

Answer 30

Question #2

Answer 31

it asked for tangent line

Answer 32

not slope

Answer 33

ops nvm

Answer 34

but yes slope is required you just need more info

Answer 35

Im derping agn, I got it now, insert into point-slope form to get tangent equation

Answer 36

ok, im going on study break now, Gonna take a step outside haha

Answer 37

Thanks all! :)

Answer 38

lol good luck

Answer 39

and sorry didn't mean to give you the answer on that one
I kind of thought they had an example or something with that one definition already given

Answer 40

Oh no problem at all, you been greatly helpful so far.