Math Problem Statement

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 =   

Solution

To analyze the function f(x)=xx2+36f(x) = \frac{x}{\sqrt{x^2 + 36}} over the interval 6x7-6 \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)=x2+361xxx2+36(x2+36)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)=(x2+36)x2(x2+36)3/2=36(x2+36)3/2f'(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)f'(x):

Using the quotient rule again:

f(x)=0(x2+36)3/23632(x2+36)1/22x(x2+36)3f''(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)=108x(x2+36)5/2f''(x) = -\frac{108x}{(x^2 + 36)^{5/2}}

Step 3: Determine Concavity

Set f(x)=0f''(x) = 0:

108x=0    x=0-108x = 0 \implies x = 0

Intervals for Concavity:

  • For x<0x < 0: f(x)>0f''(x) > 0 (concave up)
  • For x>0x > 0: f(x)<0f''(x) < 0 (concave down)

Concavity Results

  • Concave Down on: x=0x = 0 to x=7x = 7
  • Concave Up on: x=6x = -6 to x=0x = 0

Step 4: Inflection Point

The inflection point occurs where the concavity changes:

  • Inflection Point: x=0x = 0

Step 5: Find Extrema

We evaluate f(x)f(x) at the endpoints and the critical point (where f(x)=0f'(x) = 0 does not apply here):

  • f(6)=636+36=672=662=120.707f(-6) = \frac{-6}{\sqrt{36 + 36}} = \frac{-6}{\sqrt{72}} = -\frac{6}{6\sqrt{2}} = -\frac{1}{\sqrt{2}} \approx -0.707
  • f(0)=00+36=0f(0) = \frac{0}{\sqrt{0 + 36}} = 0
  • f(7)=749+36=7850.758f(7) = \frac{7}{\sqrt{49 + 36}} = \frac{7}{\sqrt{85}} \approx 0.758

Extrema Results

  • Minimum at: x=6x = -6
  • Maximum at: x=7x = 7

Summary

  • Concave Down on: 00 to 77
  • Concave Up on: 6-6 to 00
  • Inflection Point at: x=0x = 0
  • Minimum at: x=6x = -6
  • Maximum at: x=7x = 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!

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Math Problem Analysis

Mathematical Concepts

Calculus
Concavity
Inflection Points
Derivatives
Extrema

Formulas

First Derivative: f'(x) = d/dx [x / sqrt(x^2 + 36)]
Second Derivative: f''(x) for concavity and inflection points
Concavity Test: f''(x) > 0 (concave up), f''(x) < 0 (concave down)
Extrema Evaluation: Check endpoints and inflection points

Theorems

Quotient Rule for Derivatives
Concavity Test
Critical Points Evaluation

Suitable Grade Level

Grades 11-12