Math Problem Statement

Answer number 32 and 36. Show work

Answer number 32 and 36. Show work

Solution

Let's start with problem 32 and 36 from the image.

Problem 32:

f(x)=x2+3x+2f(x) = -x^2 + 3x + 2

We are tasked with finding the derivative f(x)f'(x), and then calculating f(1)f'(1), f(2)f'(2), and f(3)f'(3).

Step 1: Find the derivative

To differentiate f(x)=x2+3x+2f(x) = -x^2 + 3x + 2, we apply the power rule for each term:

  • The derivative of x2-x^2 is 2x-2x
  • The derivative of 3x3x is 33
  • The derivative of the constant 22 is 00

Thus, the derivative is: f(x)=2x+3f'(x) = -2x + 3

Step 2: Calculate f(1)f'(1), f(2)f'(2), and f(3)f'(3)

  1. f(1)=2(1)+3=2+3=1f'(1) = -2(1) + 3 = -2 + 3 = 1
  2. f(2)=2(2)+3=4+3=1f'(2) = -2(2) + 3 = -4 + 3 = -1
  3. f(3)=2(3)+3=6+3=3f'(3) = -2(3) + 3 = -6 + 3 = -3

Problem 36:

f(x)=6x2f(x) = \frac{6}{x} - 2

Step 1: Find the derivative

To differentiate f(x)=6x2f(x) = \frac{6}{x} - 2, we rewrite the term 6x\frac{6}{x} as 6x16x^{-1}:

  • The derivative of 6x16x^{-1} using the power rule is 6x2-6x^{-2}, which simplifies to 6x2-\frac{6}{x^2}
  • The derivative of 2-2 is 00

Thus, the derivative is: f(x)=6x2f'(x) = -\frac{6}{x^2}

Step 2: Calculate f(1)f'(1), f(2)f'(2), and f(3)f'(3)

  1. f(1)=612=6f'(1) = -\frac{6}{1^2} = -6
  2. f(2)=622=64=32f'(2) = -\frac{6}{2^2} = -\frac{6}{4} = -\frac{3}{2}
  3. f(3)=632=69=23f'(3) = -\frac{6}{3^2} = -\frac{6}{9} = -\frac{2}{3}

Summary of Results:

  • Problem 32:

    • f(x)=2x+3f'(x) = -2x + 3
    • f(1)=1f'(1) = 1
    • f(2)=1f'(2) = -1
    • f(3)=3f'(3) = -3
  • Problem 36:

    • f(x)=6x2f'(x) = -\frac{6}{x^2}
    • f(1)=6f'(1) = -6
    • f(2)=32f'(2) = -\frac{3}{2}
    • f(3)=23f'(3) = -\frac{2}{3}

Would you like more details on any of these steps?

Here are 5 related questions you could explore:

  1. How do you apply the power rule for derivatives in general?
  2. What is the quotient rule for derivatives, and when should it be used?
  3. How can we differentiate functions with more complex exponents?
  4. What is the significance of the derivative at specific points like f(1),f(2),f(3)f'(1), f'(2), f'(3)?
  5. How can we visualize the behavior of f(x)f(x) by examining its derivative?

Tip: The power rule is the simplest method for differentiating polynomial terms, but for more complex expressions involving division, the quotient rule or chain rule might be necessary.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Power Rule
Rational Functions

Formulas

f'(x) = d/dx [x^n] = n * x^(n-1)
f'(x) for 1/x = -1/x^2

Theorems

Power Rule
Derivative of a constant
Derivative of Rational Functions

Suitable Grade Level

Grades 11-12