Questions: Use a sign chart for f'(x) to determine the intervals on which the function is increasing or decreasing. (Express your answer using interval notation.) f(x)=x^3+4x^2+4x+3 increasing decreasing

Use a sign chart for f'(x) to determine the intervals on which the function is increasing or decreasing. (Express your answer using interval notation.) f(x)=x^3+4x^2+4x+3 increasing decreasing
Transcript text: Use a sign chart for $f^{\prime}$ to determine the intervals on which the function is increasing or decreasing. (Express your answer using interval notation.) \[ f(x)=x^{3}+4 x^{2}+4 x+3 \] increasing decreasing
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Solution

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Solution Steps

To determine the intervals on which the function \( f(x) = x^3 + 4x^2 + 4x + 3 \) is increasing or decreasing, we need to follow these steps:

  1. Find the first derivative \( f'(x) \).
  2. Determine the critical points by setting \( f'(x) = 0 \) and solving for \( x \).
  3. Use the critical points to create a sign chart for \( f'(x) \).
  4. Analyze the sign chart to determine the intervals where \( f'(x) \) is positive (increasing) and where \( f'(x) \) is negative (decreasing).
Step 1: Find the First Derivative

To determine the intervals on which the function \( f(x) = x^3 + 4x^2 + 4x + 3 \) is increasing or decreasing, we first need to find its first derivative \( f'(x) \).

\[ f(x) = x^3 + 4x^2 + 4x + 3 \]

Using the power rule for differentiation:

\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(4x^2) + \frac{d}{dx}(4x) + \frac{d}{dx}(3) \]

\[ f'(x) = 3x^2 + 8x + 4 \]

Step 2: Find the Critical Points

Next, we find the critical points by setting the first derivative equal to zero and solving for \( x \).

\[ 3x^2 + 8x + 4 = 0 \]

We solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 8 \), and \( c = 4 \).

\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \]

\[ x = \frac{-8 \pm \sqrt{64 - 48}}{6} \]

\[ x = \frac{-8 \pm \sqrt{16}}{6} \]

\[ x = \frac{-8 \pm 4}{6} \]

\[ x = \frac{-8 + 4}{6} \quad \text{or} \quad x = \frac{-8 - 4}{6} \]

\[ x = \frac{-4}{6} \quad \text{or} \quad x = \frac{-12}{6} \]

\[ x = -\frac{2}{3} \quad \text{or} \quad x = -2 \]

Step 3: Create a Sign Chart for \( f'(x) \)

We now create a sign chart for \( f'(x) = 3x^2 + 8x + 4 \) to determine the intervals on which \( f(x) \) is increasing or decreasing. We test the sign of \( f'(x) \) in the intervals determined by the critical points \( x = -2 \) and \( x = -\frac{2}{3} \).

  • For \( x < -2 \), choose \( x = -3 \): \[ f'(-3) = 3(-3)^2 + 8(-3) + 4 = 27 - 24 + 4 = 7 \quad (\text{positive}) \]

  • For \( -2 < x < -\frac{2}{3} \), choose \( x = -1 \): \[ f'(-1) = 3(-1)^2 + 8(-1) + 4 = 3 - 8 + 4 = -1 \quad (\text{negative}) \]

  • For \( x > -\frac{2}{3} \), choose \( x = 0 \): \[ f'(0) = 3(0)^2 + 8(0) + 4 = 4 \quad (\text{positive}) \]

Final Answer

Based on the sign chart:

  • The function \( f(x) \) is increasing on the intervals \( (-\infty, -2) \cup \left(-\frac{2}{3}, \infty\right) \).
  • The function \( f(x) \) is decreasing on the interval \( \left(-2, -\frac{2}{3}\right) \).

\[ \boxed{\text{Increasing: } (-\infty, -2) \cup \left(-\frac{2}{3}, \infty\right)} \] \[ \boxed{\text{Decreasing: } \left(-2, -\frac{2}{3}\right)} \]

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