Answer the following questions for the function
f(x)=x√x2+36f(x)=xx2+36
defined on the interval −6≤x≤7-6≤x≤7.

f(x)f(x) is concave down on the interval *x* =    to *x* =   

f(x)f(x) is concave up on the interval *x* =    to *x* =   

The inflection point for this function is at *x* =   

The minimum for this function occurs at *x* =   

The maximum for this function occurs at *x* =   

To analyze the function \( f(x) = \frac{x}{\sqrt{x^2 + 36}} \) over the interval \( -6 \leq x \leq 7 \), we will find its concavity, inflection points, and extrema. 

### Step 1: Find the First Derivative
Using the quotient rule:

\[
f'(x) = \frac{\sqrt{x^2 + 36} \cdot 1 - x \cdot \frac{x}{\sqrt{x^2 + 36}}}{(x^2 + 36)}
\]

Simplifying this gives:

\[
f'(x) = \frac{(x^2 + 36) - x^2}{(x^2 + 36)^{3/2}} = \frac{36}{(x^2 + 36)^{3/2}}
\]

### Step 2: Find the Second Derivative
To analyze concavity, we differentiate \( f'(x) \):

Using the quotient rule again:

\[
f''(x) = \frac{0 \cdot (x^2 + 36)^{3/2} - 36 \cdot \frac{3}{2}(x^2 + 36)^{1/2} \cdot 2x}{(x^2 + 36)^3}
\]

This simplifies to:

\[
f''(x) = -\frac{108x}{(x^2 + 36)^{5/2}}
\]

### Step 3: Determine Concavity
Set \( f''(x) = 0 \):

\[
-108x = 0 \implies x = 0
\]

**Intervals for Concavity:**
- For \( x < 0 \): \( f''(x) > 0 \) (concave up)
- For \( x > 0 \): \( f''(x) < 0 \) (concave down)

### Concavity Results
- **Concave Down on:** \( x = 0 \) to \( x = 7 \)
- **Concave Up on:** \( x = -6 \) to \( x = 0 \)

### Step 4: Inflection Point
The inflection point occurs where the concavity changes:

- **Inflection Point:** \( x = 0 \)

### Step 5: Find Extrema
We evaluate \( f(x) \) at the endpoints and the critical point (where \( f'(x) = 0 \) does not apply here):

- \( f(-6) = \frac{-6}{\sqrt{36 + 36}} = \frac{-6}{\sqrt{72}} = -\frac{6}{6\sqrt{2}} = -\frac{1}{\sqrt{2}} \approx -0.707 \)
- \( f(0) = \frac{0}{\sqrt{0 + 36}} = 0 \)
- \( f(7) = \frac{7}{\sqrt{49 + 36}} = \frac{7}{\sqrt{85}} \approx 0.758 \)

### Extrema Results
- **Minimum at:** \( x = -6 \)
- **Maximum at:** \( x = 7 \)

### Summary
- **Concave Down on:** \( 0 \) to \( 7 \)
- **Concave Up on:** \( -6 \) to \( 0 \)
- **Inflection Point at:** \( x = 0 \)
- **Minimum at:** \( x = -6 \)
- **Maximum at:** \( x = 7 \)

If you need further details or explanations, let me know! Here are some follow-up questions:

1. Would you like to see a graph of the function?
2. Do you need help with the derivation of the first and second derivatives?
3. Would you like to know about the behavior of the function beyond the interval?
4. Are you interested in how to find critical points for other functions?
5. Do you want a step-by-step breakdown of the concavity test? 

**Tip:** Always check the endpoints and critical points when looking for extrema!

Answer the following questions for the function f(x) = x / sqrt(x^2 + 36) defined on the interval -6 ≤ x ≤ 7.

f(x) is concave down on the interval x = ___ to x = ___

f(x) is concave up on the interval x = ___ to x = ___

The inflection point for this function is at x = ___

The minimum for this function occurs at x = ___

The maximum for this function occurs at x = ___

Explore the concavity, inflection points, and extrema for the function f(x) = x / sqrt(x^2 + 36) over the interval -6 ≤ x ≤ 7. This problem provides a step-by-step analysis using first and second derivatives, demonstrating concavity tests and how to locate the minimum and maximum values within the interval.

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