Rational average rate of change problem

**Devi09** · April 10th 2011, 01:51 PM

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)

**pickslides** · April 10th 2011, 02:10 PM

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?

**Devi09** · April 10th 2011, 02:15 PM

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

**pickslides** · April 10th 2011, 02:20 PM

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

**Devi09** · April 10th 2011, 02:24 PM

$-2x + 8 = 2x$

$4x=8$

$x=2$

**pickslides** · April 10th 2011, 02:41 PM

Originally Posted by Devi09

$-2x + 8 = 2x$

$4x=8$

$x=2$

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

**Devi09** · April 10th 2011, 02:55 PM

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

**pickslides** · April 10th 2011, 03:00 PM

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.

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