Zenmo
  • Zenmo
Find an equation of the tangent line to the curve at the given point (Calculus Derivatives).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Zenmo
  • Zenmo
Question #1
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freckles
  • freckles
ok what did you find for y'?
Zenmo
  • Zenmo
\[\lim_{h \rightarrow 0}\frac{ f(a+h)-f(a) }{ h }\] = \[\lim_{h \rightarrow 0}\frac{ [(3+h)^3-3(3+h)+1-(3^3-3(3)+1)] }{ h }\] . Putting more work on here second

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FireKat97
  • FireKat97
do you have to apply differentiation by first principles? :/
Zenmo
  • Zenmo
\[(3+h)^3; (3+h)(3+h)=9+3h+3h+h^2=9+6h+h^2 ; (9+6h+h^2)(3+h) = 27+18h+3h^2+9h+6h^2+h^3\]
anonymous
  • anonymous
Are you familiar with the power rule of differentiation? \[\frac{ d }{ dx }[x^n]=n*x^{n-1}\]
freckles
  • freckles
you might like pascal's triangle 4th row says 1 3 3 1 \[(3+h)^3=1(3)^3+3(3)^2h+3(3)h^2+h^3\]
Zenmo
  • Zenmo
Our class isn't on there yet, this is the beginning of calculus for the end of Chapter 2
freckles
  • freckles
\[\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h}\]
Zenmo
  • Zenmo
Sec. I might found the answer.
Zenmo
  • Zenmo
I gotten 6x+1 as the answer.
freckles
  • freckles
ok let's go slow then you do see above the 3^3 's cancel right?
freckles
  • freckles
since 3^3-3^3=0 and also you should see the 3(3) also known as 9's cancel you know since -9+9=0
Zenmo
  • Zenmo
This is how the book does it. Similar problem.
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freckles
  • freckles
you should also see the 1's cancel since 1-1=0
freckles
  • freckles
\[\lim_{h \rightarrow 0} \frac{3^3+3(3)^2h+3(3)h^2+h^3-3(3+h)+1-(3^3-3(3)+1)}{h} \\ \lim_{h \rightarrow 0} \frac{3(3)^2h+3(3)h^2+h^3-3h}{h}\]
freckles
  • freckles
now divide top and bottom by h
Zenmo
  • Zenmo
\[\lim_{h \rightarrow 0}(h^2+9h+27)\] Did u get that?
freckles
  • freckles
should wind up with \[\lim_{h \rightarrow 0} (3(3)^2+3(3)h+h^2-3)\]
freckles
  • freckles
something happened to your -3 part above
freckles
  • freckles
should be \[\lim_{h \rightarrow 0} (h^2+9h+27-3)\]
freckles
  • freckles
so your slope on your homework is off which will also make your y-intercept you found off too
freckles
  • freckles
please let me know if I need to redo the algebra for you if you don't understand
freckles
  • freckles
I guess you can call it algebra/arithmetic whatever that stuff is
Zenmo
  • Zenmo
\[\lim_{h \rightarrow 0}(h^2+9h-24)\] that correct? I'm looking at calculations at the moment.
freckles
  • freckles
well 27-3 is 24 not -24 :p
Zenmo
  • Zenmo
By using the point (3,19) and using the point-slope form for slope M=24. I get \[y=24x+19\]
freckles
  • freckles
y-19=f'(3)(x-3) y-19=24(x-3) y-19=24x-72 hmmm...-72+19...
Zenmo
  • Zenmo
\[y=24x+53\]
freckles
  • freckles
I hate to be that guy but -72+19 is -53
Zenmo
  • Zenmo
Yea so the answer is \[y=24x-53\] ?
freckles
  • freckles
yep
freckles
  • freckles
do you understand how we got the slope to be 24?
Zenmo
  • Zenmo
Yea, turns out I was messing up on simple arithmetic/algebra breaks for: adding/subtracting/factoring etc.
Zenmo
  • Zenmo
Yea I understood the core concepts, just messed up with the things from above.
Zenmo
  • Zenmo
Time for a study break haha
Zenmo
  • Zenmo
Thanks a lot for sticking around and helping. :)
freckles
  • freckles
we all mess on arithmetic just go pet a cat and then come back you will be already (that is my only study advice; cats for the win!)

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