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

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the curve defined by y = √(2x + 1) at the point (4, 3). The limit definition of the derivative is utilized, specifically M = lim (x → 4) [(√(2x + 1) - 3) / (x - 4)]. A critical error identified in the calculations was the incorrect simplification of the numerator, which should yield 2x - 8 instead of 2x - 3. Correcting this allows for straightforward evaluation of the limit, leading to the tangent line's slope and equation.

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shwanky
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Homework Statement


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

Homework Equations


a = 4
M =\lim_{x \to a} \frac{f(x) - f(a)}{x-a}

or

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

The Attempt at a Solution



M =\lim_{x \to 4} \frac{f(x) - f(4)}{x-a}

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

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

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

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

At this point I get stuck... I'm not sure what I can do from here. did I go wrong some where?
 
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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 (\sqrt{2x+1}-3) * (\sqrt{2x+1}+3) = 2x+1 -9 = 2x-8
 
Last edited:
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 \frac{0}{0} or \frac{\infty}{\infty}
 
Last edited:
BAH! thanks :).
 

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