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

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Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Here are the formulas: \[\lim_{x \rightarrow a}\frac{ f(x)f(a) }{ xa }\] (#1)

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{h \rightarrow 0}\frac{ f(a+h)f(a) }{ h } \] (#2)

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Two example problems: (A) \[y=x^33x+1; Point (3,19) \] (B) \[y=\sqrt{x} ; Point (25,5)\]

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1For 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you can use either 1 or 2

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Ah I see, so either can work.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i used to get familiar with equation 2 for all the problems when i started calc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0equation 2 is derived by equation 1. let h=xa

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its just a slightly more neater way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sure freckles will give a more clear view

freckles
 one year ago
Best ResponseYou've already chosen the best response.1example: \[\lim_{x \rightarrow 3} \frac{f(x)f(3)}{x3} \\ =\lim_{x \rightarrow 3} \frac{(x^33x+1)(3^33(3)+1)}{x3} \\ =\lim_{x \rightarrow 3} \frac{(x^33^3)+(3x+3(3))+(11)}{x3 } \\ =\lim_{x \rightarrow 3} \frac{(x3)(x^2+3x+3^2)3(x3)}{x3} \\ \text{ divide \top and bottom by } x3\] \[\lim_{x \rightarrow 3} \frac{(x^2+3x+3^2)3}{1} =\lim_{x \rightarrow 3}(x^2+3x+3^23) \\ \text{ plug in 3} \\ 3^2+3(3)+3^23=3(3^2)3=273=24\] I hope it is obvious I used the difference of cubes formula above

freckles
 one year ago
Best ResponseYou've already chosen the best response.1oops and that example was for f(x)=x^33x+1 at x=3 :

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow a}\frac{ f(x)f(a) }{ xa }\] let some variable \[h=xa\] Substitute 'h' into the first equation (note x=h+a) \[\lim_{ha \rightarrow a}\frac{ f(a+h)f(a) }{ h }\] \[\lim_{h \rightarrow 0}\frac{ f(a+h)f(a) }{ h }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0both are interrelated

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think i made a mistake

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1second, I need "checking" on this problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{h+a \rightarrow a}\frac{ f(a+h)f(a) }{ h }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{h \rightarrow 0} \frac{\sqrt{25+h}\sqrt{25}}{h} \\ \text{ multiply top and bottom by top's conjugate } \\ \lim_{h \rightarrow 0} \frac{\sqrt{25+h}\sqrt{25}}{h} \cdot \frac{\sqrt{25+h}+\sqrt{25}}{\sqrt{25+h}+\sqrt{25}} \\ \lim_{h \rightarrow 0} \frac{(25+h)(25)}{h(\sqrt{25+h}+\sqrt{25})} \\ \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{25+h}+\sqrt{25})} \\ \frac{h}{h}=1 \text{ so } \\ \lim_{h \rightarrow 0} \frac{1}{\sqrt{25+h}+\sqrt{25}}=\frac{1}{\sqrt{25}+\sqrt{25}} =\frac{1}{2 \sqrt{25}} =\frac{1}{2(5)}=\frac{1}{10}\] and this is using the other route for the other function you had there at x=25

freckles
 one year ago
Best ResponseYou've already chosen the best response.1though yeah to prove they are the same definition and you can use a substitution like @chris00 is doing above

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1If I use formula #2 for example B. \[\lim_{h \rightarrow 0}\frac{ f(25+h)f(25) }{ h }\] =\[\lim_{h \rightarrow 0}\frac{ \sqrt{25+h}\sqrt{25} }{ h } \] =\[\frac{ \sqrt{25+h}5 }{ h }\times \frac{ \sqrt{25+h}+5 }{\sqrt{25+h} +5 }\] =\[\lim_{h \rightarrow 0}\frac{ 1 }{ \sqrt{25+h}+5 }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yes!!!! replace h with 0 now

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1but for some reason, it counts that as wrong answer

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what is the question?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1it asked for tangent line

freckles
 one year ago
Best ResponseYou've already chosen the best response.1but yes slope is required you just need more info

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Im derping agn, I got it now, insert into pointslope form to get tangent equation

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1ok, im going on study break now, Gonna take a step outside haha

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and 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

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.1Oh no problem at all, you been greatly helpful so far.
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