Math Problem Statement

Use a linear approximation (or differentials) to estimate the given number. (Do not round your answer). (32.04)^4/5

Solution

To estimate (32.04)4/5(32.04)^{4/5} using a linear approximation (or differentials), we can follow these steps:

Step 1: Identify the function

We can define a function: f(x)=x4/5f(x) = x^{4/5} We are interested in estimating f(32.04)f(32.04). Since 32.04 is close to 32, we will approximate using f(32)f(32) as the base.

Step 2: Compute the base value and derivative

First, evaluate the function and its derivative at x=32x = 32.

  1. f(32)=324/5f(32) = 32^{4/5}
  2. To find the derivative of f(x)=x4/5f(x) = x^{4/5}, apply the power rule: f(x)=45x1/5f'(x) = \frac{4}{5}x^{-1/5} Now, compute the values for x=32x = 32: f(32)=324/5=16f(32) = 32^{4/5} = 16 f(32)=45321/5=4512=410=0.4f'(32) = \frac{4}{5} \cdot 32^{-1/5} = \frac{4}{5} \cdot \frac{1}{2} = \frac{4}{10} = 0.4

Step 3: Apply the linear approximation formula

The linear approximation formula is: f(x)f(a)+f(a)(xa)f(x) \approx f(a) + f'(a)(x - a) where a=32a = 32 and x=32.04x = 32.04.

Substitute the values: f(32.04)f(32)+f(32)(32.0432)f(32.04) \approx f(32) + f'(32)(32.04 - 32) f(32.04)16+0.4(32.0432)f(32.04) \approx 16 + 0.4 \cdot (32.04 - 32) f(32.04)16+0.40.04f(32.04) \approx 16 + 0.4 \cdot 0.04 f(32.04)16+0.016f(32.04) \approx 16 + 0.016 f(32.04)16.016f(32.04) \approx 16.016

Thus, the linear approximation for (32.04)4/5(32.04)^{4/5} is 16.01616.016.

Would you like more details on any step?

Here are five questions that relate to this topic:

  1. How does the power rule help in finding the derivative for functions with fractional exponents?
  2. Why is using differentials effective for estimating small changes in functions?
  3. How could you apply this method to a function with a higher-degree root, like (50.03)3/7(50.03)^{3/7}?
  4. What are the limits of using linear approximations for larger deviations from the base point?
  5. Can this technique be extended to estimate non-polynomial functions like logarithms?

Tip: Always select a base point close to the value you are estimating to improve the accuracy of the linear approximation.

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Math Problem Analysis

Mathematical Concepts

Linear Approximation
Differentials
Exponents

Formulas

f(x) ≈ f(a) + f'(a)(x - a)
f'(x) = (4/5)x^(-1/5)

Theorems

Power Rule for Differentiation

Suitable Grade Level

Grades 11-12