Question

Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)

Answer 1

Answer:

$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$

First tangent line:

$y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5$

Second tangent line:

$y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7$

Notice: slope of -1 means that both $f'(x_1), \ f'(x_2)$ are equal to -1, so $f'(x_1)=-1 \ and \ f'(x_2)=-1$

Answer 2

Answer: y = -x + 5   and    y = -x - 7     (see attached graph)

Step-by-step explanation:

y = $\frac{9}{x + 1}$

  = 9(x + 1)⁻¹

Use the product rule to find the derivative

a = 9           a' = 0

b = (x + 1)⁻¹   b' = -(x + 1)⁻²

 ab' + a'b

= 9[-(x + 1)⁻²] + 0[(x + 1)⁻¹ ]

= $\frac{-9}{(x + 1)^{2}}$

Set the derivative equal to the desired slope of -1 to solve for x

-1 = $\frac{-9}{(x + 1)^{2}}$

-(x + 1)² = -9

 (x + 1)² = 9

 √(x + 1)² = √9  

    x + 1 = +/- 3

x + 1 = 3      x + 1 = -3

     x = 2          x = -4

Plug those values into the original equation to solve for y:

y = $\frac{9}{x + 1}$

  = $\frac{9}{2 + 1}$

  = 3

(2, 3)

y = $\frac{9}{x + 1}$

  = $\frac{9}{-4 + 1}$

  = -3

(-4, -3)

Next, plug in the given slope (-1) and the coordinates above into the Point-Slope formula y - y₁ = m(x - x₁) to find the equations:

m = -1, (x₁ y₁) = (2, 3)                             m = -1, (x₁ y₁) = (-4, -3)

y - 3 = -1(x - 2)                                       y + 3 = -1(x + 4)

y - 3 = -x + 2                                          y + 3 = -x - 4

    y = -x + 5                                                y = -x - 7