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Find the equation of a tangent line to y=f(x)

  • Thread starter shwanky
  • Start date
  • #1
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Homework Statement


Find the equation of the tangent line to the curve at the given point.
[tex]y = \sqrt{2x+1} , (4, 3)[/tex]


Homework Equations


a = 4
[tex]M =\lim_{x \to a} \frac{f(x) - f(a)}{x-a}[/tex]

or

[tex]Mpq = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}[/tex]




The Attempt at a Solution



[tex]M =\lim_{x \to 4} \frac{f(x) - f(4)}{x-a}[/tex]

[tex]M =\lim_{x \to 4} \frac{\sqrt{2x+1} - \sqrt{2 * 4 + 1}}{x-4}[/tex]

[tex]M =\lim_{x \to 4} \frac{\sqrt{2x+1} - 3}{x-4}[/tex]

[tex]M =\lim_{x \to 4} \frac{\sqrt{2x+1} - 3}{x-4} * \frac{\sqrt{2x+1} + 3}{\sqrt{2x+1} + 3}[/tex]

[tex]M =\lim_{x \to 4} \frac{2x - 3}{x-4(\sqrt{2x+1} + 3)}[/tex]

At this point I get stuck... I'm not sure what I can do from here. did I go wrong some where?
 

Answers and Replies

  • #2
On the last step are you sure that you multiplied correctly? :P (check the numerator again)

Do you see it?

It should be 2x-8 since [tex](\sqrt{2x+1}-3) * (\sqrt{2x+1}+3) = 2x+1 -9 = 2x-8 [/tex]
 
Last edited:
  • #3
From that point you can divide by (x-4) quite easily resulting in a function for which you can directly substitute x=4 since the result is not in an 'indeterminate form' of [tex]\frac{0}{0}[/tex] or [tex]\frac{\infty}{\infty} [/tex]
 
Last edited:
  • #4
43
0
BAH! thanks :).
 

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