# Rational average rate of change problem

• Apr 10th 2011, 02:51 PM
Devi09
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)
• Apr 10th 2011, 03:10 PM
pickslides
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?
• Apr 10th 2011, 03:15 PM
Devi09
m= (4 - 10)/(8-5)
m= -2
• Apr 10th 2011, 03:20 PM
pickslides
So now solve $\displaystyle -2 = \left(\frac{2x}{x-4}\right)'$
• Apr 10th 2011, 03:24 PM
Devi09
$-2x + 8 = 2x$

$4x=8$

$x=2$
• Apr 10th 2011, 03:41 PM
pickslides
Quote:

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

$4x=8$

$x=2$

Nope, $\displaystyle \left(\frac{2x}{x-4}\right)'= \left(\frac{2(x-4)-2x\times 1}{(x-4)^2}\right)$
• Apr 10th 2011, 03:55 PM
Devi09
Quote:

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
• Apr 10th 2011, 04:00 PM
pickslides
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.