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:See explanationStep-by-step explanation:You are given the equation of the curve[tex]y=\dfrac{x}{1+x}[/tex]Point [tex]P\left(1,\dfrac{1}{2}\right)[/tex] lies on the curve.Point [tex]Q\left(x,\dfrac{x}{1+x}\right)[/tex] is an arbitrary point on the curve.The slope of the secant line PQ is[tex]\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1][/tex]1. If x=0.5, then the slope is[tex]\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333[/tex]2. If x=0.9, then the slope is[tex]\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632[/tex]3. If x=0.99, then the slope is[tex]\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513[/tex]4. If x=0.999, then the slope is[tex]\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501[/tex]5. If x=1.5, then the slope is[tex]\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2[/tex]6. If x=1.1, then the slope is[tex]\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381[/tex]7. If x=1.01, then the slope is[tex]\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488[/tex]8. If x=1.001, then the slope is[tex]\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499[/tex]