f(x) = x/x^2 + 10x + 24 find increasing decreasing of f

To find the intervals where the function \( f(x) = \frac{x}{x^2 + 10x + 24} \) is increasing or decreasing, we need to follow these steps:

### 1. Find the derivative of \( f(x) \)
Using the quotient rule for differentiation, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then:
\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}
\]
Here, \( g(x) = x \) and \( h(x) = x^2 + 10x + 24 \).

First, find the derivatives of \( g(x) \) and \( h(x) \):
- \( g'(x) = 1 \)
- \( h'(x) = 2x + 10 \)

Now, apply the quotient rule:
\[
f'(x) = \frac{(1)(x^2 + 10x + 24) - (x)(2x + 10)}{(x^2 + 10x + 24)^2}
\]

Simplify the numerator:
\[
f'(x) = \frac{x^2 + 10x + 24 - (2x^2 + 10x)}{(x^2 + 10x + 24)^2}
\]
\[
f'(x) = \frac{x^2 + 10x + 24 - 2x^2 - 10x}{(x^2 + 10x + 24)^2}
\]
\[
f'(x) = \frac{-x^2 + 24}{(x^2 + 10x + 24)^2}
\]

### 2. Determine where \( f'(x) = 0 \) or undefined
Set the numerator of \( f'(x) \) equal to zero:
\[
-x^2 + 24 = 0
\]
\[
x^2 = 24
\]
\[
x = \pm \sqrt{24} = \pm 2\sqrt{6}
\]

### 3. Analyze the sign of \( f'(x) \)
We need to determine the sign of \( f'(x) \) in different intervals, which are determined by the critical points \( x = -2\sqrt{6} \) and \( x = 2\sqrt{6} \).

We also need to check when the denominator \( (x^2 + 10x + 24)^2 \) is zero. The quadratic equation \( x^2 + 10x + 24 = 0 \) has no real solutions (since its discriminant is negative). Therefore, the function is never undefined for real values of \( x \).

The critical points divide the number line into three intervals:
- \( (-\infty, -2\sqrt{6}) \)
- \( (-2\sqrt{6}, 2\sqrt{6}) \)
- \( (2\sqrt{6}, \infty) \)

Now we can test the sign of \( f'(x) \) in each of these intervals by plugging test points into the derivative \( f'(x) = \frac{-x^2 + 24}{(x^2 + 10x + 24)^2} \).

### 4. Conclusion on increasing and decreasing intervals
- **Increasing:** Where \( f'(x) > 0 \)
- **Decreasing:** Where \( f'(x) < 0 \)

By testing points from each interval, we find:
- The function is **increasing** on the interval \( (-2\sqrt{6}, 2\sqrt{6}) \).
- The function is **decreasing** on the intervals \( (-\infty, -2\sqrt{6}) \) and \( (2\sqrt{6}, \infty) \).

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Would you like more details on the analysis of the sign of the derivative?

### Related questions:
1. How do we find critical points of more complex functions?
2. What are other methods to determine increasing and decreasing intervals?
3. How does the quotient rule work for different types of functions?
4. How do concavity and inflection points relate to increasing and decreasing intervals?
5. How do second derivatives help in determining the nature of a function?

### Tip:
When applying the quotient rule, always simplify the numerator and denominator carefully to avoid calculation errors, especially in complex functions.

Learn how to find the intervals of increase and decrease for the function f(x) = x/(x^2 + 10x + 24) using the quotient rule. This solution involves differentiation, finding critical points, and analyzing intervals. Suitable for advanced high school or undergraduate students.

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