Question

El punto P(1,1/2) está sobre la curva y=x/(1+x).
(a) Si Q es el punto (x,x/(1+x)), use su calculadora para hallar la pendiente de la recta secante PQ (correcta hasta seis cifras decimales) para los valores de x que se enumeran a continuación:
(i) 0.5 (ii) 0.9 (iii) 0.99 (iv) 0.999 (v) 1.5 (vi) 1.1 (vii) 1.01 (viii) 1.001
(b) Mediante los resultados del inciso (a) conjeture el valor de la pendiente de la recta tangente a la curva en P (1,1/2).
(c) Usando la pendiente del inciso (b) encuentre la ecuación de la recta tangente a la curva en P (1,1/2).

Answer 1

Answer:

Teacher is god of the polynomial f the same time as the one jdjg

Step-by-step explanation:

0.5

Answer 2

Answer:

The Slope of PQ line according to given values of x are:

(i). 0.333, (ii). 0.27, (iii). 0.3, (iv). 1, (v). 0.2, (vi). 0.23, (vii). 0.2, (viii). 0.2

Step-by-step explanation:

The equation of the curve is $y = \frac{x}{x+1}$ (Given).

And point P(1, 1/2) lie on the given curve.

Given the another point is Q($x,\frac{x}{x+1}$).

(i). If we put x = 0.5, then

     $\frac{x}{x+1} = \frac{0.5}{0.5+1} =\frac{1}{3}$

    After putting the value, the point Q is (0.5, 1/3).

    So, the slope of PQ line = $\frac{y_{2} - y_{1} }{x_{2}-x_{1} }$

                                             = $\frac{\frac{1}{3}-\frac{1}{2} }{0.5 - 1}$

                                             = 0.333

(ii). Similarly, at x = 0.9, $\frac{x}{x+1} = \frac{0.9}{0.9+1} = 0.473$

    Then point Q is ( 0.9, 0.473).

     So, the slope of PQ line = $\frac{0.473 - \frac{1}{2} }{0.9 - 1} = 0.27$

(iii). At x= 0.99, $\frac{x}{x+1} = \frac{0.99}{0.99 +1} = 0.497$

      Then point Q is (0.99, 0.497).

     So, the slope of PQ line = $\frac{0.497 - \frac{1}{2} }{0.99 - 1} = 0.3$

(iv). At x = 0.999, $\frac{x}{x +1} = \frac{0.999}{0.999+1} = 0.499$

     Then point Q is (0.999, 0.499).

     So, the Slope of PQ line = $\frac{0.499 - \frac{1}{2} }{0.999-1} = 1$

(v). At x = 1.5, $\frac{x}{x+1} = \frac{1.5}{1.5 +1} = 0.6$

     Then point Q is ( 1.5, 0.6).

     So, the slope of PQ line = $\frac{0.6 - \frac{1}{2} }{1.5 -1} = 0.2$

(vi). At x = 1.1, $\frac{x}{x+1} = \frac{1.1}{1.1+1} = 0.523$

      Then point Q is (1.1, 0.523).

      So, the slope of PQ line = $\frac{0.523 - \frac{1}{2} }{1.1 - 1}$$= 0.23$

(vii). At x= 1.01, $\frac{x}{x+1} = \frac{1.01}{1.01+1} = 0.502$

      Then point Q is (1.01, 0.502).

      So, the slope of PQ line = $\frac{0.502 - \frac{1}{2} }{1.01 -1} = 0.2$

(viii). At x = 1.001, $\frac{x}{x+1} = \frac{1.001}{1.001 +1} = 0.5002$

       Then point Q is (1.001, 0.5002).

       So, the slope of PQ line = $\frac{0.5002 - \frac{1}{2} }{1.001 -1} = 0.2$