the relative maximum value of the function h(x)=20x/3x^(2)+20 is ? Note: round off your answer to the nearest hundredths.

To find the relative maximum value of the function \( h(x) = \frac{20x}{3x^2 + 20} \), we follow these steps:

### Step 1: Compute the first derivative
We need to find the critical points by taking the first derivative of the function \( h(x) \).

Using the quotient rule for differentiation, which is:

\[
\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}
\]

Here, \( f(x) = 20x \) and \( g(x) = 3x^2 + 20 \).

- \( f'(x) = 20 \)
- \( g'(x) = 6x \)

Now, applying the quotient rule:

\[
h'(x) = \frac{20(3x^2 + 20) - 20x(6x)}{(3x^2 + 20)^2}
\]

Simplify the numerator:

\[
h'(x) = \frac{20(3x^2 + 20) - 120x^2}{(3x^2 + 20)^2} = \frac{60x^2 + 400 - 120x^2}{(3x^2 + 20)^2}
\]

\[
h'(x) = \frac{-60x^2 + 400}{(3x^2 + 20)^2}
\]

### Step 2: Set \( h'(x) = 0 \)
To find critical points, we set the derivative equal to zero:

\[
\frac{-60x^2 + 400}{(3x^2 + 20)^2} = 0
\]

This implies that:

\[
-60x^2 + 400 = 0
\]

Solving for \( x \):

\[
60x^2 = 400 \implies x^2 = \frac{400}{60} = \frac{20}{3}
\]

\[
x = \pm \sqrt{\frac{20}{3}} \approx \pm 2.58
\]

### Step 3: Determine if the critical points are maxima or minima
To determine whether these critical points are relative maxima or minima, we compute the second derivative or check the behavior of the first derivative around these points.

However, for now, we evaluate the function \( h(x) \) at the critical points \( x \approx 2.58 \) and \( x \approx -2.58 \).

### Step 4: Evaluate \( h(x) \) at \( x \approx 2.58 \)
Substitute \( x = 2.58 \) into the original function:

\[
h(2.58) = \frac{20(2.58)}{3(2.58)^2 + 20} \approx \frac{20(2.58)}{3(6.6564) + 20} = \frac{51.6}{39.9692} \approx 1.29
\]

Thus, the relative maximum value is approximately **1.29**.

### Final Answer:
The relative maximum value of the function is approximately **1.29**.

Do you want more details or have any questions?

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Here are 5 related questions for further practice:
1. How do you find the relative minimum value of a function?
2. What is the role of the second derivative in determining concavity?
3. How can you use the first derivative to identify points of inflection?
4. How does the quotient rule apply in real-world situations?
5. What is the significance of critical points in optimizing functions?

**Tip:** The second derivative test can quickly tell whether a critical point is a local maximum or minimum based on its sign at that point.

The relative maximum value of the function h(x) = 20x / (3x^2 + 20) is? Note: round off your answer to the nearest hundredths.

Discover how to find the relative maximum value of the function h(x) = 20x / (3x^2 + 20) by using the first derivative and critical points. This solution involves calculus concepts such as the quotient rule and derivative tests, making it ideal for students in Grades 11-12 or introductory calculus courses.

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