Use a calculator to evaluate the function at the indicated values. Round your answers to three decimals.
h(x) = e−3x;    h

1
3

, h(1.5), h(−1), h(−𝜋)
h

1
3

=	  
Incorrect: Your answer is incorrect.
h(1.5)
=	 
h(−1)
=	 
h(−𝜋)
=	 

import numpy as np

# Define the function h(x) = e^(-3x)
def h(x):
    return np.exp(-3 * x)

# Calculate h(x) for the given values
values = {
    "h(1/3)": h(1/3),
    "h(1.5)": h(1.5),
    "h(-1)": h(-1),
    "h(-π)": h(-np.pi)
}

# Round the results to three decimal places
rounded_values = {key: round(val, 3) for key, val in values.items()}
rounded_values{'h(1/3)': 0.368, 'h(1.5)': 0.011, 'h(-1)': 20.086, 'h(-π)': 12391.648}{'h(1/3)': 0.368, 'h(1.5)': 0.011, 'h(-1)': 20.086, 'h(-π)': 12391.648}successHere are the rounded values for the function \( h(x) = e^{-3x} \):

1. \( h\left(\frac{1}{3}\right) = 0.368 \)
2. \( h(1.5) = 0.011 \)
3. \( h(-1) = 20.086 \)
4. \( h(-\pi) = 12391.648 \)

Do you want further explanation or details on how these values were computed?

Here are 5 related questions to expand your understanding:

1. How does the value of \( e^{-3x} \) change as \( x \) becomes more positive or negative?
2. What is the general effect of a negative exponent on the behavior of exponential functions?
3. How does the base \( e \) influence the properties of \( e^{-3x} \) compared to other exponential bases?
4. How would the graph of \( h(x) = e^{-3x} \) look for \( x \in [-\pi, 2] \)?
5. Can the function \( h(x) \) ever reach a value of 0? Why or why not?

**Tip:** When rounding intermediate results for exponential functions, always ensure the precision aligns with the context of the problem!

Use a calculator to evaluate the function at the indicated values. Round your answers to three decimals. h(x) = e^(-3x); h(1/3), h(1.5), h(-1), h(-π)

Learn how to evaluate the exponential function h(x) = e^(-3x) at given values, including h(1/3), h(1.5), h(-1), and h(-π). The step-by-step solution explains rounding to three decimals and explores properties of exponential functions. Suitable for students in Grades 10-12.

Math Problem Statement

= Incorrect: Your answer is incorrect. h(1.5)

h(−1)

h(−𝜋)

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