Math Problem Statement

find the relative maximum and minimum

Find the relative maximum and minimum for the given functions.

Solution

To find the relative maximum and minimum of the given functions, we need to perform the following steps for each function:

  1. Find the derivative of the function.
  2. Set the derivative equal to zero and solve for xx. These values of xx are the critical points.
  3. Determine the nature of each critical point (whether it's a maximum, minimum, or saddle point) using the second derivative test or by analyzing the sign changes in the first derivative.

Let's begin by solving for the first few functions from the image you provided. I'll start by finding the derivative of each function and then we will solve for the critical points.

Function 22: f(x)=3x36x2+8f(x) = 3x^3 - 6x^2 + 8

  1. First derivative: f(x)=9x212xf'(x) = 9x^2 - 12x

  2. Set the derivative to zero and solve for xx: 9x212x=09x^2 - 12x = 0 Factor out 3x3x: 3x(3x4)=03x(3x - 4) = 0 This gives us x=0x = 0 and x=43x = \frac{4}{3}.

  3. Second derivative: f(x)=18x12f''(x) = 18x - 12

    • At x=0x = 0, f(0)=12f''(0) = -12, so this is a relative maximum.
    • At x=43x = \frac{4}{3}, f(43)=12f''\left(\frac{4}{3}\right) = 12, so this is a relative minimum.

Function 23: g(x)=2x3+7x5g(x) = -2x^3 + 7x - 5

  1. First derivative: g(x)=6x2+7g'(x) = -6x^2 + 7
  2. Set the derivative to zero and solve for xx: \quad \Rightarrow \quad x = \pm \sqrt{\frac{7}{6}} \approx \pm 1.08$$
  3. Second derivative: g(x)=12xg''(x) = -12x
    • At x=76x = \sqrt{\frac{7}{6}}, g(x)<0g''(x) < 0, indicating a relative maximum.
    • At x=76x = -\sqrt{\frac{7}{6}}, g(x)>0g''(x) > 0, indicating a relative minimum.

Function 24: f(x)=x4+3x32f(x) = -x^4 + 3x^3 - 2

  1. First derivative: f(x)=4x3+9x2f'(x) = -4x^3 + 9x^2
  2. Set the derivative to zero and solve for xx: \quad \Rightarrow \quad x^2(-4x + 9) = 0 \quad \Rightarrow \quad x = 0, \, x = \frac{9}{4} = 2.25$$
  3. Second derivative: f(x)=12x2+18xf''(x) = -12x^2 + 18x
    • At x=0x = 0, f(0)=0f''(0) = 0 (use other methods like sign analysis).
    • At x=2.25x = 2.25, f(2.25)=54f''(2.25) = -54, so this is a relative maximum.

Let me know if you would like me to continue with more functions or clarify any step!


Here are 5 related questions:

  1. What is the importance of the second derivative test in finding extrema?
  2. Can a function have both absolute and relative extrema at the same point?
  3. How do we determine if a critical point is a saddle point?
  4. What happens if the second derivative is zero at a critical point?
  5. What are the roles of concavity in determining the shape of the graph near critical points?

Tip: Always double-check your critical points by plugging them back into the original function to understand their nature and ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Calculus
Critical Points
Maxima and Minima
Derivatives
Second Derivative Test

Formulas

First Derivative: f'(x)
Second Derivative: f''(x)
Second Derivative Test: If f''(x) > 0, it's a local minimum. If f''(x) < 0, it's a local maximum.

Theorems

Second Derivative Test
Critical Point Theorem

Suitable Grade Level

Grades 11-12, College Calculus