Question

1. $f(x)=\frac{x^{2}}{4x + 4}$

ap increasing:
ho decreasing:

Explanation:

Step1: Find the derivative

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u = x^{2}\), \(u^\prime=2x\); \(v = 4x + 4\), \(v^\prime=4\). Then \(f^\prime(x)=\frac{2x(4x + 4)-x^{2}\times4}{(4x + 4)^{2}}=\frac{8x^{2}+8x-4x^{2}}{(4x + 4)^{2}}=\frac{4x^{2}+8x}{(4x + 4)^{2}}=\frac{4x(x + 2)}{16(x + 1)^{2}}=\frac{x(x + 2)}{4(x + 1)^{2}}\)

Step2: Find critical points

Set \(f^\prime(x)=0\), then \(x(x + 2)=0\), so \(x = 0\) or \(x=-2\). The function \(y = f(x)\) is not defined at \(x=-1\) (since \(v=4x + 4 = 0\) when \(x=-1\))

Step3: Test intervals

• For the interval \((-\infty,-2)\), pick \(x=-3\). \(f^\prime(-3)=\frac{-3(-3 + 2)}{4(-3+1)^{2}}=\frac{-3\times(-1)}{4\times4}=\frac{3}{16}>0\)
• For the interval \((-2,-1)\), pick \(x =-\frac{3}{2}\). \(f^\prime(-\frac{3}{2})=\frac{-\frac{3}{2}(-\frac{3}{2}+ 2)}{4(-\frac{3}{2}+1)^{2}}=\frac{-\frac{3}{2}\times\frac{1}{2}}{4\times(-\frac{1}{2})^{2}}=\frac{-\frac{3}{4}}{4\times\frac{1}{4}}=-\frac{3}{4}<0\)
• For the interval \((-1,0)\), pick \(x=-\frac{1}{2}\). \(f^\prime(-\frac{1}{2})=\frac{-\frac{1}{2}(-\frac{1}{2}+ 2)}{4(-\frac{1}{2}+1)^{2}}=\frac{-\frac{1}{2}\times\frac{3}{2}}{4\times\frac{1}{4}}=-\frac{3}{4}<0\)
• For the interval \((0,\infty)\), pick \(x = 1\). \(f^\prime(1)=\frac{1(1 + 2)}{4(1 + 1)^{2}}=\frac{3}{16}>0\)

Answer:

Increasing: \((-\infty,-2)\cup(0,\infty)\)
Decreasing: \((-2,-1)\cup(-1,0)\)