How Do You Find the Point Where a Tangent Equals a Secant Slope?

[eleventhxhour]

Given the function f(x) = 3x / (x - 2), determine the coordinates of a point on f(x) for 3 < x < 6 where the slope of the tangent line is equal to the slope of the secant line passing through A(3, 9) and B(6, 9/2).

So I found that the slope of the secant line is -1.5 (therefore slope of tangent is also -1.5) and then I'm not sure what to do next. I draw a graph of the function, and got that the answer should be around 3.2-3.5 for the x-coordinate but the answer in the book has it as (4, 6). How would you find this exact answer (algebraically)?

Thanks

 

[MarkFL]

Since you posted this in the Pre-Calculus forum, I would suggest looking at the difference quotient of the given function, and take the limit as $h\to0$. THis will give you the slope as a function of $x$, and would in fact be the derivative with respect to $x$.

So, begin with:

\(\displaystyle f'(x)\equiv\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\)

Now, use the given function $f$, plug into the formula above, and then simplify the expression algebraically and the let $h\to0$. Then equate the resulting derivative, or $f'(x)$, to the secant slope of \(\displaystyle -\frac{3}{2}\), and solve for $x$. You will get a quadratic in $x$, where only one root is in the given interval.