<p>Please help me!! .. Im stuck</p>
<p>Find the point (s) on the graph of y = 1/x where the tangent line is parallel to 4x + 9y = 3</p>
<p>Thank you = )</p>
<p>anyone know how to do this??? Please!!</p>
<p>The best place for non-SAT/ACT calculus questions is here:</p>
<h1><a href="http://physicsforums.com/forumdisplay.php?f=156%5B/url%5D">http://physicsforums.com/forumdisplay.php?f=156</a></h1>
<p>What's the slope of the line tangent to the graph of y=f(x) in point (x, y)?</p>
<p>What the connection between slopes of parallel lines?</p>
<p>After you find x of the point of tangency, plug it into y=1/x to get y.</p>
<p>You might get more then one point.
What could that mean?</p>
<p>What is the derivative of 1/x? Also known as x^-1...</p>
<p>f'(x) = -x^-2, or -1/(x^2)</p>
<p>put 4x + 9y = 3 in a format where you can see the slope:
9y = 3 - 4x
y = 1/3 - (4/9)x
so the slope of this line is -4/9. Any lines parallel to it will also have that slope.
So, where is f'(x) = -4/9?</p>
<p>-1/(x^2) = -4/9
1/(x^2) = 4/9
x^2 = 9/4
x = +/- 3/2
and there is your answer: {-1.5, +1.5}</p>
<p>Have a nice day.</p>
<p>thank you very much!</p>