Answer:
[tex]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[/tex]
First tangent line:
[tex]y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5[/tex]
Second tangent line:
[tex]y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7[/tex]
Notice: slope of -1 means that both [tex]f'(x_1), \ f'(x_2)[/tex] are equal to -1, so [tex]f'(x_1)=-1 \ and \ f'(x_2)=-1[/tex]
Answer: y = -x + 5 and y = -x - 7 (see attached graph)
Step-by-step explanation:
y = [tex]\frac{9}{x + 1}[/tex]
= 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 ]
= [tex]\frac{-9}{(x + 1)^{2}}[/tex]
Set the derivative equal to the desired slope of -1 to solve for x
-1 = [tex]\frac{-9}{(x + 1)^{2}}[/tex]
-(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 = [tex]\frac{9}{x + 1}[/tex]
= [tex]\frac{9}{2 + 1}[/tex]
= 3
(2, 3)
y = [tex]\frac{9}{x + 1}[/tex]
= [tex]\frac{9}{-4 + 1}[/tex]
= -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
