Answer
The function is increasing on \( (-\infty, -3) \) and \( (3, \infty) \).
Solution
To determine the intervals on which the function \( f(x) = x^3 - 27x + 4 \) is increasing or decreasing, as well as to find the local extrema, we will follow these steps:
1. **Find the derivative \( f'(x) \)** to determine the critical points.
2. **Set the derivative equal to zero** to find the critical points.
3. **Analyze the sign of the derivative** to determine where the function is increasing or decreasing.
4. **Identify the local extrema** based on the critical points.
Let's start by calculating the derivative of \( f(x) \):
\[
f'(x) = \frac{d}{dx}(x^3 - 27x + 4)
\]
Now, I will calculate the derivative.
Simplify the expression by following steps:
- step0: Solution:
\(3x^{2}-27\)
The derivative of the function is:
\[
f'(x) = 3x^2 - 27
\]
Next, we will set the derivative equal to zero to find the critical points:
\[
3x^2 - 27 = 0
\]
Now, I will solve this equation for \( x \).
Solve the quadratic equation by following steps:
- step0: Solve using square roots:
\(3x^{2}-27=0\)
- step1: Add or subtract both sides:
\(3x^{2}=27\)
- step2: Divide both sides:
\(\frac{3x^{2}}{3}=\frac{27}{3}\)
- step3: Divide the numbers:
\(x^{2}=9\)
- step4: Simplify the expression:
\(x=\pm \sqrt{9}\)
- step5: Simplify the expression:
\(x=\pm 3\)
- step6: Separate into possible cases:
\(\begin{align}&x=3\\&x=-3\end{align}\)
- step7: Rewrite:
\(x_{1}=-3,x_{2}=3\)
The critical points are:
\[
x_1 = -3 \quad \text{and} \quad x_2 = 3
\]
Next, we will analyze the sign of the derivative \( f'(x) = 3x^2 - 27 \) in the intervals determined by the critical points: \( (-\infty, -3) \), \( (-3, 3) \), and \( (3, \infty) \).
1. **Choose test points** in each interval:
- For \( (-\infty, -3) \), choose \( x = -4 \)
- For \( (-3, 3) \), choose \( x = 0 \)
- For \( (3, \infty) \), choose \( x = 4 \)
2. **Evaluate the sign of \( f'(x) \)** at these test points.
Let's calculate \( f'(-4) \), \( f'(0) \), and \( f'(4) \).
Calculate the value by following steps:
- step0: Calculate:
\(3\times 0^{2}-27\)
- step1: Calculate:
\(3\times 0-27\)
- step2: Multiply the numbers:
\(0-27\)
- step3: Remove 0:
\(-27\)
Calculate or simplify the expression \( 3*(4)^2 - 27 \).
Calculate the value by following steps:
- step0: Calculate:
\(3\times 4^{2}-27\)
- step1: Multiply the terms:
\(48-27\)
- step2: Subtract the numbers:
\(21\)
Calculate or simplify the expression \( 3*(-4)^2 - 27 \).
Calculate the value by following steps:
- step0: Calculate:
\(3\left(-4\right)^{2}-27\)
- step1: Multiply the terms:
\(48-27\)
- step2: Subtract the numbers:
\(21\)
The evaluations of the derivative at the test points are as follows:
- \( f'(-4) = 21 \) (positive)
- \( f'(0) = -27 \) (negative)
- \( f'(4) = 21 \) (positive)
Now we can summarize the behavior of the function \( f(x) \):
- On the interval \( (-\infty, -3) \), \( f'(x) > 0 \) (increasing).
- On the interval \( (-3, 3) \), \( f'(x) < 0 \) (decreasing).
- On the interval \( (3, \infty) \), \( f'(x) > 0 \) (increasing).
### Summary of Intervals:
- The function is increasing on \( (-\infty, -3) \) and \( (3, \infty) \).
- The function is decreasing on \( (-3, 3) \).
### Local Extrema:
- At \( x = -3 \), there is a local maximum (since the function changes from increasing to decreasing).
- At \( x = 3 \), there is a local minimum (since the function changes from decreasing to increasing).
Now, let's summarize the answers:
A. The function is increasing on:
\[
(-\infty, -3) \cup (3, \infty)
\]
B. The function is never increasing: **False** (the function is increasing in the specified intervals).
Thus, the final answer is:
A. The function is increasing on \( (-\infty, -3) \cup (3, \infty) \).
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