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

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To find the point where the tangent slope equals the secant slope for the function f(x) = 3x / (x - 2) between 3 < x < 6, first calculate the secant slope between points A(3, 9) and B(6, 9/2), which is -1.5. The next step involves using the difference quotient to derive the function's derivative, f'(x), by taking the limit as h approaches 0. Set the derivative equal to the secant slope of -1.5 and solve the resulting quadratic equation for x. The solution will yield a valid x-coordinate within the specified interval, which is confirmed to be around 4, aligning with the book's answer.
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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
 
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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:

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

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 $$-\frac{3}{2}$$, and solve for $x$. You will get a quadratic in $x$, where only one root is in the given interval.
 
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