Math Problem Statement

In Exercises 21–24, use differentials and the graph of f to approximate (a) f(1.9) and (b) f(2.04).

Solution

To solve these problems, we will use the concept of differentials to approximate f(1.9)f(1.9) and f(2.04)f(2.04). Here's how to approach it:

General Steps:

  1. The tangent line to the curve at (2,1)(2, 1) is used to approximate values of f(x)f(x) near x=2x = 2.
  2. The tangent line equation is derived using: yf(2)=f(2)(x2)y - f(2) = f'(2)(x - 2) Here, f(2)f'(2) is the slope of the tangent line at x=2x = 2, which can be determined from the graph.
  3. Use the tangent line to approximate the function value.

Problem 21 (a): Approximating f(1.9)f(1.9):

  • The point on the curve is (2,1)(2, 1), so f(2)=1f(2) = 1.
  • From the graph, the slope f(2)1020=0.5f'(2) \approx \frac{1 - 0}{2 - 0} = 0.5 (approximated visually from the graph).
  • The equation of the tangent line becomes: y1=0.5(x2)y=0.5x1y - 1 = 0.5(x - 2) \quad \Rightarrow \quad y = 0.5x - 1
  • For x=1.9x = 1.9: f(1.9)0.5(1.9)1=0.951=0.05f(1.9) \approx 0.5(1.9) - 1 = 0.95 - 1 = -0.05

(b) Approximating f(2.04)f(2.04):

  • Using the same tangent line equation y=0.5x1y = 0.5x - 1: f(2.04)0.5(2.04)1=1.021=0.02f(2.04) \approx 0.5(2.04) - 1 = 1.02 - 1 = 0.02

Problem 22 (a): Approximating f(1.9)f(1.9):

  • The point on the curve is (2,1)(2, 1), so f(2)=1f(2) = 1.
  • From the graph, the slope f(2)1.5f'(2) \approx -1.5 (approximated visually from the graph).
  • The equation of the tangent line becomes: y1=1.5(x2)y=1.5x+4y - 1 = -1.5(x - 2) \quad \Rightarrow \quad y = -1.5x + 4
  • For x=1.9x = 1.9: f(1.9)1.5(1.9)+4=2.85+4=1.15f(1.9) \approx -1.5(1.9) + 4 = -2.85 + 4 = 1.15

(b) Approximating f(2.04)f(2.04):

  • Using the same tangent line equation y=1.5x+4y = -1.5x + 4: f(2.04)1.5(2.04)+4=3.06+4=0.94f(2.04) \approx -1.5(2.04) + 4 = -3.06 + 4 = 0.94

Final Results:

  • Problem 21:
    • f(1.9)0.05f(1.9) \approx -0.05
    • f(2.04)0.02f(2.04) \approx 0.02
  • Problem 22:
    • f(1.9)1.15f(1.9) \approx 1.15
    • f(2.04)0.94f(2.04) \approx 0.94

Would you like more details or clarification on any step? Here are related questions:

  1. How is the slope f(x)f'(x) estimated from the graph?
  2. Why is the tangent line useful for approximations near x=2x = 2?
  3. How would the approximations change if the slope were miscalculated?
  4. Can you explain the concept of differentials in this context?
  5. How do these approximations relate to Taylor series expansions?

Tip: Always confirm the slope visually or numerically to improve accuracy when approximating from graphs.

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Math Problem Analysis

Mathematical Concepts

Differentials
Tangent Line Approximation
Functions

Formulas

y - f(a) = f'(a)(x - a)
Linear Approximation Formula: f(x) ≈ f(a) + f'(a)(x - a)

Theorems

Linear Approximation Theorem

Suitable Grade Level

High School (Grades 11-12) or early college