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Find the area of the triangle formed by the x-axis, the y-axis, and the tangent line to the graph of $y = \frac{1}{\sqrt{x}}$ at the point where $x=k$.

I have no clue how to do this and my teacher never explained area to us... Where should I start? I found the derivative of $y = \frac{1}{\sqrt{x}}$ but that's all I have right now. Thank you.

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The next step is to write the equation of the tangent line.

If $x=k$, then $y=\frac{1}{\sqrt{k}}$, so $(k, \frac{1}{\sqrt{k}})$ is on the line. The slope of the line needs to match the derivative at $x=k$, so $m = -\frac{1}{2} k^{-3/2}$.

A line through a known point and with known slope can always be written:

$$ y-y_0 = m(x-x_0) $$

so in this case:

$$ y - \frac{1}{\sqrt{k}} = -\frac{1}{2} k^{-3/2} (x-k) .$$

Then you will need to find where that line intersects the $x$-axis ($y=0$) and the $y$-axis ($x=0$). For area of that triangle, just use the usual triangle formula $A=\frac{1}{2}bh$.

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