Verified answer

y = 9x/(x + 1) ← this is a curve → in red

The function looks like (u/v), so its derivative looks like: [(u'.v) - (v'.u)]/v² → where:

u = 9x → u' = 9

v = x + 1 → v' = 1

y' = [(u'.v) - (v'.u)]/v²

y' = [9.(x + 1) - 9x]/(x + 1)²

y' = (9x + 9 - 9x)/(x + 1)²

y' = 9/(x + 1)² ← this is the derivative

…but the derivative is too the slope of the tangent line to the curve at x

x - y = 0 ← this is a line → in blue

- y = - x

y = x ← the slope is 1

Two lines are parallel if they have the same slope. So the tangent line to the curve must have a slope of 1 to be parallel to the line of equation: x - y = 0

To find when the tangent to the curve has a slope of 1 is similar to solve the equation:

y' = 1

9/(x + 1)² = 1

9 = (x + 1)²

9 - (x + 1)² = 0

3² - (x + 1)² = 0 → you recognize: a² - b² = (a + b).(a - b)

[3 + (x + 1)].[3 - (x + 1)] = 0

(3 + x + 1).(3 - x - 1) = 0

(x + 4).(- x + 2) = 0

First case: (x + 4) = 0 → x + 4 = 0 → x = - 4

y = 9x/(x + 1) → when: x = - 4

y = - 36/- 3 = 12

→ first point A (- 4 ; 12)

Second case: (- x + 2) = 0 → - x + 2 = 0 → - x = - 2 → x = 2

y = 9x/(x + 1) → when: x = 2

y = 18/3 = 6

→ second point B (2 ; 6)

Equation of the straight line

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept

The slope of the line is 1, so the equation of the line becomes: y = x + b

The first line passes through A, so the coordinates of this point must verify the equation of the line.

y = x + b

b = y - x → you substitute x and y by the coordinates of the point A (- 4 ; 12)

b = 12 + 4 = 16

→ The equation of the first tangent line to the curve is: y = x + 16 → in green

The second line passes through B, so the coordinates of this point must verify the equation of the line.

y = x + b

b = y - x → you substitute x and y by the coordinates of the point B (2 ; 6)

b = 6 + 2 = 4

→ The equation of the first tangent line to the curve is: y = x + 4 → in PURPLE

Donde la recta tangente sea horizontal?

When a line is horizontal, the slope is null.

To find when the tangent to the curve has a slope of 0 is similar to solve the equation:

y' = 0

9/(x + 1)² = 0 → no possible

Hola

y = 9 x / (x + 1)

ln(y) = ln(9) + ln(x) - ln(x + 1)

derivamos

y'/y = (1/x) - (1/(x + 1)) = ((x + 1) - x) / (x (x + 1))

y'= y * ( 1 / (x (x + 1)) )

y' = (9 x / (x + 1)) (1 / (x (x + 1)))

Pendiente de la tangente (derivada)

y' = 9 / (x + 1)^2

********************

a)

Para que la tangente sea paralela a

y - x = 0

y = x

con pendiente

m = 1

se debe cumplir

y' = 9 / (x + 1)^2 = 1

entonces

(x + 1)^2 = 9

Dos soluciones

Primera

x1 + 1 = -3

x1 = -3 - 1

x1 = -4

**********

Segunda

x2 + 1 = 3

x2 = 3 - 1

x2 = 2

************

b)

Para que sea horizontal la tangente

y' = 9 / (x + 1)^2 = 0

Este resultado sólo se consigue

cuando x crece/decrece indefinidamente

es decir, para x tendiendo a infinito

a) Tangentes a la curva y = (9x/ x + 1 )que sean paralelas a la recta x -y = 0

Derivando, obtenemos la pendiente:

y' = [9.(x+1) - 9x] / (x+1)^2

y' = 9 / (x+1)^2

La pendiente de x-y=0 es: --> y=x --> pendiente m=1

Igualando:

9 / (x+1)^2 = 1

9 = x^2 + 2x +1

x^2 + 2x - 8 = 0

Resolviéndola, da x = 2, x=-4

Sustituyendo en la ecuación de la curva:

y = 9x / (x+1) ,

con x=2 --> y = 18/3 --> y = 6 --> punto de tangencia (2, 6)

con x=-4 --> y = -36 / -3 --> y = 12 --> punto (-4, 12)

La ecuación de la tangente paralela a y = x es y=x+k

con (2, 6) --> 6=2+k --> k=4 --> tangente y = x+4

con (-4, 12) --> 12 = -4+k --> k= 16 --> tangente y = x + 16

b) La horizontal tiene pendiente m= cero

La pendiente de la tangente era 9 / (x+1)^2

para que sea igual a cero, el numerador 9 debería ser cero.

No hay tangente horizontal