Math Problem Statement

Analyze and solve the problem involving the function f(x) = -2x^3 + 12x^2 + 30x, focusing on finding its maximum value, derivative, and critical points within the interval [1, 6].

Solution

Let’s analyze and solve the questions from the document.

Given:

The function for power P(n)P(n) is modeled as: f(x)=2x3+12x2+30xf(x) = -2x^3 + 12x^2 + 30x where xx represents the frequency of rotation in thousands of revolutions per minute (rpm), within the interval [1,6][1, 6].


1.1.b. For which value of xx does the function f(x)f(x) appear to have a maximum?

To determine the maximum:

  1. Find the derivative f(x)f'(x).
  2. Solve f(x)=0f'(x) = 0 to locate critical points.
  3. Test the critical points within [1,6][1, 6] to identify the maximum.

Derivative:

f(x)=ddx(2x3+12x2+30x)f'(x) = \frac{d}{dx}(-2x^3 + 12x^2 + 30x) f(x)=6x2+24x+30f'(x) = -6x^2 + 24x + 30

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

6x2+24x+30=0-6x^2 + 24x + 30 = 0 Simplify: x24x5=0x^2 - 4x - 5 = 0 Factorize: (x5)(x+1)=0(x - 5)(x + 1) = 0 x=5orx=1x = 5 \quad \text{or} \quad x = -1 Since xx must be within the interval [1,6][1, 6], only x=5x = 5 is valid.

Thus, the maximum appears at x=5x = 5.


1.2.a. Expression for the derivative f(x)f'(x):

From earlier: f(x)=6x2+24x+30f'(x) = -6x^2 + 24x + 30


1.2.b. Calculate f(5)f'(5):

Substitute x=5x = 5 into f(x)f'(x): f(5)=6(5)2+24(5)+30f'(5) = -6(5)^2 + 24(5) + 30 f(5)=6(25)+120+30f'(5) = -6(25) + 120 + 30 f(5)=150+120+30=0f'(5) = -150 + 120 + 30 = 0

This confirms x=5x = 5 is indeed a critical point.


Would you like me to graph the function or provide more detailed steps? Let me know! Here are five additional questions based on this:

  1. How do you confirm whether x=5x = 5 is a maximum or a minimum?
  2. What is the value of f(x)f(x) at x=5x = 5?
  3. How can the second derivative test be applied to verify the nature of the critical point?
  4. What happens to f(x)f(x) as x6x \to 6?
  5. Can f(x)f(x) have another critical point within the interval [1,6][1, 6]?

Tip:

Always verify critical points by substituting into the second derivative f(x)f''(x) or testing intervals around them to confirm maxima or minima.

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Math Problem Analysis

Mathematical Concepts

Polynomial functions
Critical points
Optimization

Formulas

Derivative of f(x)
Critical points: f'(x) = 0
Factoring quadratic equations

Theorems

Test for maxima and minima
Quadratic factorization

Suitable Grade Level

Grades 10-12