Zenmo
  • Zenmo
Small Question (formula related). Finding equation of tangent line at the given point.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Zenmo
  • Zenmo
Here are the formulas: \[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\] (#1)
Zenmo
  • Zenmo
\[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h } \] (#2)
Zenmo
  • Zenmo
Two example problems: (A) \[y=x^3-3x+1; Point (3,19) \] (B) \[y=\sqrt{x} ; Point (25,5)\]

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More answers

Zenmo
  • Zenmo
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?
freckles
  • freckles
you can use either 1 or 2
anonymous
  • anonymous
yea..
Zenmo
  • Zenmo
Ah I see, so either can work.
anonymous
  • anonymous
i used to get familiar with equation 2 for all the problems when i started calc
anonymous
  • anonymous
equation 2 is derived by equation 1. let h=x-a
anonymous
  • anonymous
its just a slightly more neater way
anonymous
  • anonymous
I'm sure freckles will give a more clear view
freckles
  • freckles
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
freckles
  • freckles
oops and that example was for f(x)=x^3-3x+1 at x=3 :
anonymous
  • anonymous
\[\lim_{x \rightarrow a}\frac{ f(x)-f(a) }{ x-a }\] let some variable \[h=x-a\] Substitute 'h' into the first equation (note x=h+a) \[\lim_{h-a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }\] \[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\]
anonymous
  • anonymous
both are inter-related
anonymous
  • anonymous
oops
anonymous
  • anonymous
i think i made a mistake
Zenmo
  • Zenmo
second, I need "checking" on this problem.
anonymous
  • anonymous
should be h+a==>a
anonymous
  • anonymous
\[\lim_{h+a \rightarrow a}\frac{ f(a+h)-f(a) }{ h }\]
freckles
  • freckles
\[\lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \\ \text{ multiply top and bottom by top's conjugate } \\ \lim_{h \rightarrow 0} \frac{\sqrt{25+h}-\sqrt{25}}{h} \cdot \frac{\sqrt{25+h}+\sqrt{25}}{\sqrt{25+h}+\sqrt{25}} \\ \lim_{h \rightarrow 0} \frac{(25+h)-(25)}{h(\sqrt{25+h}+\sqrt{25})} \\ \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{25+h}+\sqrt{25})} \\ \frac{h}{h}=1 \text{ so } \\ \lim_{h \rightarrow 0} \frac{1}{\sqrt{25+h}+\sqrt{25}}=\frac{1}{\sqrt{25}+\sqrt{25}} =\frac{1}{2 \sqrt{25}} =\frac{1}{2(5)}=\frac{1}{10}\] and this is using the other route for the other function you had there at x=25
freckles
  • freckles
though yeah to prove they are the same definition and you can use a substitution like @chris00 is doing above
Zenmo
  • Zenmo
If I use formula #2 for example B. \[\lim_{h \rightarrow 0}\frac{ f(25+h)-f(25) }{ h }\] =\[\lim_{h \rightarrow 0}\frac{ \sqrt{25+h}-\sqrt{25} }{ h } \] =\[\frac{ \sqrt{25+h}-5 }{ h }\times \frac{ \sqrt{25+h}+5 }{\sqrt{25+h} +5 }\] =\[\lim_{h \rightarrow 0}\frac{ 1 }{ \sqrt{25+h}+5 }\]
freckles
  • freckles
yes!!!! replace h with 0 now
Zenmo
  • Zenmo
plug in 0 to get 1/10
freckles
  • freckles
yep yep
Zenmo
  • Zenmo
but for some reason, it counts that as wrong answer
freckles
  • freckles
what is the question?
freckles
  • freckles
exactly
Zenmo
  • Zenmo
1 Attachment
Zenmo
  • Zenmo
Question #2
freckles
  • freckles
it asked for tangent line
freckles
  • freckles
not slope
Zenmo
  • Zenmo
ops nvm
freckles
  • freckles
but yes slope is required you just need more info
Zenmo
  • Zenmo
Im derping agn, I got it now, insert into point-slope form to get tangent equation
Zenmo
  • Zenmo
ok, im going on study break now, Gonna take a step outside haha
Zenmo
  • Zenmo
Thanks all! :)
freckles
  • freckles
lol good luck
freckles
  • freckles
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
Zenmo
  • Zenmo
Oh no problem at all, you been greatly helpful so far.

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