For the curve y = ln(x 2 − 3), find the equation of the tangent line at (2, 0).

Mathematics

1 answer:

denis23 [38]4 years ago

3 0

Y' = 2x/(x^2 -3) y'(2) = 4
Using the point slope equation:
y-0 = 4(x-2)
y = 4x - 8

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
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\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
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\\ \quad \\
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\end{array}\\\\
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\\\\
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with that template in mind, let's see $\bf f(x)=\sqrt{x}\qquad 
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so B went form 1 on f(x), down to 0.5 or 1/2 on g(x)
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