Answer:
Increasing: (-∞, -2) ∪ (0, ∞)
Decreasing: (-2, -1) ∪ (-1, 0)
Constant: x = 0, x = -2
Step-by-step explanation:
A function is increasing when the gradient is positive ⇒ f'(x) > 0
A function is decreasing when the gradient is negative ⇒ f'(x) < 0
A function is constant when the gradient is zero ⇒ f'(x) = 0
Given function:
[tex]f(x)=\dfrac{x^2+2x+2}{x+1}[/tex]
Differentiate the given function using the quotient rule.
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Quotient Rule for Differentiation}\\\\If $y=\dfrac{u}{v}$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{v \dfrac{\text{d}u}{\text{d}x}-u\dfrac{\text{d}v}{\text{d}x}}{v^2}$\\\end{minipage}}[/tex]
[tex]\textsf{Let}\;\;u=x^2+2x+2 \implies \dfrac{\text{d}u}{\text{d}x}=2x+2[/tex]
[tex]\textsf{Let}\;\;v=x+1 \implies \dfrac{\text{d}v}{\text{d}x}=1[/tex]
Therefore:
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=\dfrac{(x+1) (2x+2)-(x^2+2x+2)}{(x+1)^2}[/tex]
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=\dfrac{2x^2+4x+2-x^2-2x-2}{(x+1)^2}[/tex]
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=\dfrac{x^2+2x}{(x+1)^2}[/tex]
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=\dfrac{x(x+2)}{(x+1)^2}[/tex]
[tex]\implies f'(x)=\dfrac{x(x+2)}{(x+1)^2}[/tex]
Find the critical values of the differentiated function (the zeros of the numerator and the denominator):
- Zeros of the numerator: x = 0, x = -2
- Zeros of the denominator: x = -1
Therefore, the intervals are:
- x < -2
- -2 < x < -1
- -1 < x < 0
- x > 0
Choose numbers that are within each interval and substitute them into the differentiated function:
[tex]x < -2 \implies f'(-3)=\dfrac{-3(-3+2)}{(-3+1)^2}=\dfrac{3}{4} > 0[/tex]
[tex]-2 < x < -1 \implies f'(-1.5)=\dfrac{-1.5(-1.5+2)}{(-1.5+1)^2}=-3 < 0[/tex]
[tex]-1 < x < 0 \implies f'(-0.5)=\dfrac{-0.5(-0.5+2)}{(-0.5+1)^2}=-3 < 0[/tex]
[tex]x > 0 \implies f'(1)=\dfrac{1(1+2)}{(1+1)^2}=\dfrac{3}{4} > 0[/tex]
Increasing
Therefore, f'(x) > 0 when:
- x < -2 : (-∞, -2)
- x > 0 : (0, ∞)
Decreasing
Therefore, f'(x) < 0 when:
- -2 < x < -1 : (-2, -1)
- -1 < x < 0 : (-1, 0)
Constant
To find the interval where f(x) is constant, set the differentiated function to zero:
[tex]\implies \dfrac{x(x+2)}{(x+1)^2} =0[/tex]
[tex]\implies x(x+2) =0[/tex]
[tex]\implies x=0 \;\; \textsf{or}\;\;x=-2[/tex]
