The point P(1,1/2) lies on the curve y=x/(1+x). If Q is the point (x , x/(1+x), how do you find slope of the secant line PQ if the values are
(1) 0.5
(2) 0.99
(3) 0.9
(4) 0.999
2006-08-23 12:59:11 · 4 answers · asked by buttrefly007 1 in Science & Mathematics ➔ Mathematics
Are those values of x at point Q? I'm confused.
2006-08-23 13:05:20 · answer #1 · answered by trueblue88 5 · 0⤊ 0⤋
The slope of the line PQ is given as
slope = (y1-y2)/(x1-x2) = ( (x/1+x)-1/2 )/ (x-1)
= ( (2x-1-x)/2(1+x) ) / (x-1)
= 1/2(1+x).
Put various values of x in this equation to get the required slopes. For example, if x=0.5, slope = 1/2*1.5 = 1/3.
I am afraid the answer given by HAI is wrong. It gives the slope of the tangent at any point x.
2006-08-25 00:16:37 · answer #2 · answered by muten 2 · 0⤊ 0⤋
1. derive the equation f(x)=x/(x+1)
2. from this resuult, f'(x)=1/(1+x)^2, substitute the values in x and you will get the slopes at those values of x.
ex. at x=0.5 the slope is 4/9
2006-08-23 20:28:33 · answer #3 · answered by hai 2 · 0⤊ 1⤋
Plug each of those values in for x, to get the 2nd point
1. x = 0.5
Q(0.5, 0.5/(1+0.5)) = Q(0.5, 0.3333)
Now just used the formula for slope to find the answer:
(y1 - y2)/(x1 - x2)
2006-08-23 20:07:52 · answer #4 · answered by godmike 2 · 0⤊ 0⤋