Find the derivative of the function h, defined for all x
by the formula h(x) = 8x/(3x^2+4) Note that h(x) →
0 as x → ±∞. Use the sign variation of h’ (x) to find
the extreme points of h(x).

We are given the function \( h(x) = \frac{8x}{3x^2 + 4} \). The task is to:

1. Find the derivative \( h'(x) \).
2. Use the sign variation of \( h'(x) \) to determine the extreme points of \( h(x) \).

### Step 1: Differentiate \( h(x) \)
The function \( h(x) \) is a quotient, so we will apply the **quotient rule**. The quotient rule is:

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

Here, \( f(x) = 8x \) and \( g(x) = 3x^2 + 4 \).

#### Derivatives of \( f(x) \) and \( g(x) \):
- \( f'(x) = 8 \)
- \( g'(x) = 6x \)

Now, apply the quotient rule:

\[
h'(x) = \frac{(8)(3x^2 + 4) - (8x)(6x)}{(3x^2 + 4)^2}
\]

Simplifying the numerator:

\[
h'(x) = \frac{24x^2 + 32 - 48x^2}{(3x^2 + 4)^2}
\]

\[
h'(x) = \frac{-24x^2 + 32}{(3x^2 + 4)^2}
\]

Thus, the derivative is:

\[
h'(x) = \frac{32 - 24x^2}{(3x^2 + 4)^2}
\]

### Step 2: Find the critical points (extreme points)
The critical points occur where \( h'(x) = 0 \), meaning the numerator of \( h'(x) \) must be zero:

\[
32 - 24x^2 = 0
\]

Solving for \( x \):

\[
24x^2 = 32
\]
\[
x^2 = \frac{32}{24} = \frac{4}{3}
\]
\[
x = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}
\]

Thus, the critical points are \( x = \pm \frac{2\sqrt{3}}{3} \).

### Step 3: Determine the sign of \( h'(x) \)
To analyze the sign of \( h'(x) = \frac{32 - 24x^2}{(3x^2 + 4)^2} \), we need to consider the numerator \( 32 - 24x^2 \):

- If \( x^2 < \frac{4}{3} \), then \( 32 - 24x^2 > 0 \), so \( h'(x) > 0 \).
- If \( x^2 > \frac{4}{3} \), then \( 32 - 24x^2 < 0 \), so \( h'(x) < 0 \).

Thus:
- \( h'(x) > 0 \) for \( -\frac{2\sqrt{3}}{3} < x < \frac{2\sqrt{3}}{3} \)
- \( h'(x) < 0 \) for \( x < -\frac{2\sqrt{3}}{3} \) and \( x > \frac{2\sqrt{3}}{3} \)

### Step 4: Conclusion about extreme points
Since \( h'(x) \) changes sign from positive to negative at \( x = \pm \frac{2\sqrt{3}}{3} \), these are **local maxima**. Therefore, the extreme points of \( h(x) \) are at \( x = \pm \frac{2\sqrt{3}}{3} \).

Would you like any further details or clarifications?

Here are 5 related questions to explore further:

1. How do you determine concavity using the second derivative of \( h(x) \)?
2. Can we find the exact value of \( h(x) \) at the critical points?
3. How does the behavior of \( h(x) \) as \( x \to \pm \infty \) relate to the existence of horizontal asymptotes?
4. What is the graphical interpretation of the quotient rule applied here?
5. Can we explore any symmetry in the function \( h(x) \) based on its form?

**Tip:** To check for local minima or maxima, always examine the sign of the first derivative around the critical points.

Find the derivative of the function h, defined for all x by the formula h(x) = 8x/(3x^2+4). Note that h(x) → 0 as x → ±∞. Use the sign variation of h’(x) to find the extreme points of h(x).

Learn how to find the derivative of the function h(x) = 8x/(3x^2 + 4) using the quotient rule. This detailed step-by-step solution explains how to determine critical points and analyze the sign variation of h'(x) to identify local maxima. Suitable for students in undergraduate calculus.

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