Math Problem Statement

Evaluate the function \( f(x) = 300e^{(0.02x)} \) at \( x = 48 \), rounding the result to three decimal places.

Solution

To solve this problem, we need to evaluate the function f(x)=300e(0.02x)f(x) = 300e^{(0.02x)} at x=48x = 48, and then round the answer to three decimal places.

Steps to solve:

  1. Substitute x=48x = 48: f(48)=300e(0.02×48)f(48) = 300e^{(0.02 \times 48)}

  2. Calculate the exponent: 0.02×48=0.960.02 \times 48 = 0.96

    So, we now have: f(48)=300e0.96f(48) = 300e^{0.96}

  3. Evaluate e0.96e^{0.96} (using the value of e2.718e \approx 2.718): e0.962.6117e^{0.96} \approx 2.6117

  4. Multiply by 300: f(48)300×2.6117=783.51f(48) \approx 300 \times 2.6117 = 783.51

  5. Round to three decimal places: f(48)783.510f(48) \approx 783.510

Final Answer:

f(48)=783.510f(48) = 783.510

Would you like a more detailed explanation or have any questions?


Here are some related questions to deepen understanding:

  1. What would f(x)f(x) equal if x=50x = 50 instead of 48?
  2. How does changing the coefficient 300 affect the function's value?
  3. What is the impact of changing the exponent base ee to a different value?
  4. How can we interpret the significance of e0.02xe^{0.02x} in real-world applications?
  5. What happens to f(x)f(x) as xx approaches infinity?

Tip: Always double-check exponent calculations, as small errors in exponent values can significantly impact the result.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Evaluation of Exponential Expressions
Rounding

Formulas

f(x) = 300e^{(0.02x)}

Theorems

Properties of Exponential Functions

Suitable Grade Level

Grades 9-12