Thread: Rational average rate of change problem

1. Rational average rate of change problem

Given the function $f(x) = \frac{\ 2x}{x-4}$ , determine the coordinates of a point on f(x) where the slope of the tangent line equals the slope of the secant line that passes through A(5,10) and B(8,4)

2. First find the slope of the secant line that passes through A(5,10) and B(8,4)

Hint: $\displaystyle m = \frac{y_2-y_1}{x_2-x_1}$

What do you get?

3. m= (4 - 10)/(8-5)
m= -2

4. So now solve $\displaystyle -2 = \left(\frac{2x}{x-4}\right)'$

5. $-2x + 8 = 2x$

$4x=8$

$x=2$

6. Originally Posted by Devi09
$-2x + 8 = 2x$

$4x=8$

$x=2$
Nope, $\displaystyle \left(\frac{2x}{x-4}\right)'= \left(\frac{2(x-4)-2x\times 1}{(x-4)^2}\right)$

7. Originally Posted by pickslides
Nope, $\displaystyle \left(\frac{2x}{x-4}\right)'= \left(\frac{2(x-4)-2x\times 1}{(x-4)^2}\right)$
How did you get that?

when you cross multiply $-2= (2x)/(x-4)$

you get x = 2

8. You need to find $\displaystyle \left(\frac{2x}{x-4}\right)'$ which is the gradient function of $\displaystyle \frac{2x}{x-4}$

$\displaystyle \left(\frac{2x}{x-4}\right)'= \left(\frac{2(x-4)-2x\times 1}{(x-4)^2}\right)$ via the quotient rule for differentiation.