anonymous
  • anonymous
find slope of tangent line y=f(x)=xsquared + 1 at (2,5) given f'(x)=2x
Mathematics
schrodinger
  • schrodinger
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cwrw238
  • cwrw238
the slope = 2x at (2,5) x = 2 so slope of tangent = 2(2) = 4
2bornot2b
  • 2bornot2b
So your question is to find the slope of the equation\(f(x)=x^2+1\) at (2,5) given f'(x)=2x
2bornot2b
  • 2bornot2b
joycelefebvre are you new to OS?

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anonymous
  • anonymous
yes could you show me your work on that please i'm studying for praxis 0061 my study guied doesnt show work its horrible tu
2bornot2b
  • 2bornot2b
OK let me help you
2bornot2b
  • 2bornot2b
Given any function, the derivative of the function is the equation to the slope of the given equation--------did you know that?
anonymous
  • anonymous
kinda I took calc in college in 93 and its not coming back
2bornot2b
  • 2bornot2b
OK, then I am trying to refresh that. Suppose you have a function y=f(x), so the function y=f'(x) is the equation of the tangent of the function y=f(x)
2bornot2b
  • 2bornot2b
OK?
2bornot2b
  • 2bornot2b
So your given function is \(f(x)=x^2+1\) and \(f'(x)=2x\)
2bornot2b
  • 2bornot2b
Now as I told you earlier, \(f'(x)\) should be the equation of the tangent at any point
2bornot2b
  • 2bornot2b
Your question asks you to find the equation of the tangent at (2,5) so just put x=2 in the equation to tangent i.e. 2x2=4. So the answer is 4 Got it?
anonymous
  • anonymous
omg I was trying to do y2-y1/x2-x1
2bornot2b
  • 2bornot2b
This is actually a theorem.. If you just remember it you don't have to find the tangent that way.......... I mean the way you are trying using the formula \(\frac {y2-y1}{x2-x1}\)
2bornot2b
  • 2bornot2b
The theorem is, Let y=f(x) be a given equation, to find the equation to the tangent at any point is y=f'(x) where f'(x) is the derivative of the function f(x)
2bornot2b
  • 2bornot2b
Is that clear, any problem?
anonymous
  • anonymous
tu i couldn't find the praxis study guied for 0061 so i got the cliff notes and its all text no examples
2bornot2b
  • 2bornot2b
Did you know you can chat with other users in the maths group?
2bornot2b
  • 2bornot2b
In that way you can get help from experts?
2bornot2b
  • 2bornot2b
Come join me in the chat room (do you know how?)
anonymous
  • anonymous
no
2bornot2b
  • 2bornot2b
Do you see those green boxes below
anonymous
  • anonymous
your 31
2bornot2b
  • 2bornot2b
See there I have posted you a chat message
2bornot2b
  • 2bornot2b
Yes, I am 31, thats my level.
2bornot2b
  • 2bornot2b
But chatting is a different thing
anonymous
  • anonymous
my lap top keeps telling me i cant run this script
2bornot2b
  • 2bornot2b
OH, I see, what operating system are you using?

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