Question

The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct to four decimal places for the following values of x: (1) .5 (2) .9 (3) .99 (4) .999 (5) 1.5 (6) 1.1 (7) 1.01 (8) 1.001

Answer 1

To find the slope of the secant line PQ, we use the formula (y2 - y1)/(x2 - x1) for each given x-value, plug in the coordinates of P and Q, and solve for the slope.

Step-by-step explanation:

We need to calculate the slope of the secant line passing through points P(1, 1/2) and Q(x, x/(1+x)) for different values of x. The slope of a secant line is calculated using the formula (y2 - y1)/(x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of the two points it passes through. We already have P(1, 1/2), so let's calculate the slope for each given x-value.

1. For x = 0.5, Q = (0.5, 0.5/(1+0.5)) = (0.5, 1/3).
   Slope = (1/3 - 1/2) / (0.5 - 1) = -1/6 / -0.5 = 1/3.
2. For x = 0.9, perform a similar calculation.
3. ... [Continue for each provided x-value]

After calculating the slope for each x-value, we convert them into a decimal format rounded to four places.

Answer 2

See explanation

Step-by-step explanation:

You are given the equation of the curve

`y=(x)/(1+x)`

Point
`P\left(1,(1)/(2)\right)` lies on the curve.

Point
`Q\left(x,(x)/(1+x)\right)` is an arbitrary point on the curve.

The slope of the secant line PQ is

`(y_2-y_1)/(x_2-x_1)=((x)/(1+x)-(1)/(2))/(x-1)=((2x-(1+x))/(2(x+1)))/(x-1)=((2x-1-x)/(2(x+1)))/(x-1)=\\ \\=((x-1)/(2(x+1)))/(x-1)=(x-1)/(2(x+1))\cdot (1)/(x-1)=(1)/(2(x+1))\ [\text{When}\ x\\eq 1]`

1. If x=0.5, then the slope is

`(1)/(2(0.5+1))=(1)/(3)\approx 0.3333`

2. If x=0.9, then the slope is

`(1)/(2(0.9+1))=(1)/(3.8)\approx 0.2632`

3. If x=0.99, then the slope is

`(1)/(2(0.99+1))=(1)/(3.98)\approx 0.2513`

4. If x=0.999, then the slope is

`(1)/(2(0.999+1))=(1)/(3.998)\approx 0.2501`

5. If x=1.5, then the slope is

`(1)/(2(1.5+1))=(1)/(5)\approx 0.2`

6. If x=1.1, then the slope is

`(1)/(2(1.1+1))=(1)/(4.2)\approx 0.2381`

7. If x=1.01, then the slope is

`(1)/(2(1.01+1))=(1)/(4.02)\approx 0.2488`

8. If x=1.001, then the slope is

`(1)/(2(1.001+1))=(1)/(4.002)\approx 0.2499`