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Write a. the sum of the two functions. b. the difference of the first function minus the second function. c. the product of the two functions. d. the quotient of the first function divided by the second function. Evaluate each of these constructed functions at 2. $$ q(t)=4 e^{2 t} ; r(t)=2\left(0.3^{t}\right) $$

Short Answer

Expert verified
Sum: \( 4e^4 + 0.18 \); Difference: \( 4e^4 - 0.18 \); Product: \( 0.72e^4 \); Quotient: \( 22.22e^4 \).

Step by step solution

01

Define the Functions

The exercise provides two functions: \[ q(t) = 4e^{2t} \] and \[ r(t) = 2(0.3)^t \] These functions will be used to find their sum, difference, product, and quotient.
02

Find the Sum of the Functions

The sum of the functions \( q(t) \) and \( r(t) \) is given by:\[ s(t) = q(t) + r(t) = 4e^{2t} + 2(0.3)^t \]To evaluate \( s(t) \) at \( t = 2 \), substitute 2 for \( t \): \[ s(2) = 4e^{4} + 2(0.3)^2 \]Calculate the values:\[ s(2) = 4e^{4} + 2(0.09) = 4e^4 + 0.18 \] (numerical evaluation of \( e \) is needed).
03

Find the Difference of the Functions

The difference of the functions \( q(t) \) and \( r(t) \) is:\[ d(t) = q(t) - r(t) = 4e^{2t} - 2(0.3)^t \]Evaluate \( d(t) \) at \( t = 2 \): \[ d(2) = 4e^{4} - 2(0.09) \]Calculate the values:\[ d(2) = 4e^4 - 0.18 \] (numerical evaluation of \( e \) is needed).
04

Find the Product of the Functions

The product of the functions \( q(t) \) and \( r(t) \) is:\[ p(t) = q(t) \times r(t) = 4e^{2t} \times 2(0.3)^t = 8e^{2t} \times (0.3)^t \]To find \( p(t) \) at \( t = 2 \): \[ p(2) = 8e^{4} \times (0.3)^2 \]Calculate the values:\[ p(2) = 8e^4 \times 0.09 = 0.72e^4 \] (numerical evaluation of \( e \) is needed).
05

Find the Quotient of the Functions

The quotient of the functions \( q(t) \) and \( r(t) \) is:\[ q(t) = \frac{q(t)}{r(t)} = \frac{4e^{2t}}{2(0.3)^t} = \frac{2e^{2t}}{0.3^t} \]Evaluate \( q(t) \) at \( t = 2 \):\[ q(2) = \frac{2e^{4}}{0.3^2} \]Calculate the value:\[ q(2) = \frac{2e^4}{0.09} \approx 22.22e^4 \] (numerical evaluation of \( e \) is needed).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are mathematical expressions in which the variable appears in the exponent. They have the general form \( f(x) = a^x \), where \( a \) is a constant, known as the base of the exponential function. In the exercise, we encounter exponential functions like \( q(t) = 4e^{2t} \) and \( r(t) = 2(0.3)^t \). Here, the bases \( e \) and \( 0.3 \) are raised to powers involving the variable \( t \). The constant \( e \) is a special mathematical constant approximately equal to 2.71828, known for its natural properties in growth processes.
  • Exponential Growth: If the base is greater than 1 (like \( e \)), the function represents exponential growth as \( t \) increases.
  • Exponential Decay: If the base is between 0 and 1 (like \( 0.3 \)), the function represents exponential decay over time.
Exponential functions model various real-world processes, including population growth, radioactive decay, and interest calculations.
Function Operations
Function operations involve methods for combining two or more functions to create a new function. These operations can be thought of like arithmetic operations but applied to functions. The main operations are:
  • Sum: Adding two functions together. For example, the sum of \( q(t) = 4e^{2t} \) and \( r(t) = 2(0.3)^t \) is \( s(t) = 4e^{2t} + 2(0.3)^t \).
  • Difference: Subtracting one function from another. Here, it's \( d(t) = 4e^{2t} - 2(0.3)^t \).
  • Product: Multiplying two functions. In this problem, \( p(t) = 4e^{2t} \times 2(0.3)^t = 8e^{2t} \times (0.3)^t \).
  • Quotient: Dividing one function by another. This is represented as \( q(t) = \frac{4e^{2t}}{2(0.3)^t} = \frac{2e^{2t}}{0.3^t} \).
Function operations allow us to craft new functions and investigate how they interact in various scenarios, providing insights across different disciplines.
Evaluating Functions
Evaluating functions involves finding the value of the function for a specific input, often to understand its behavior at certain points. It's like plugging a number into an equation to see what comes out. For instance, when we evaluate an exponential function like \( q(t) = 4e^{2t} \) at \( t = 2 \), we're asking what the function's output is when \( t \) is 2. We substitute 2 into the equation:- For the sum \( s(t) \), \( s(2) = 4e^{4} + 0.18 \).- For the difference \( d(t) \), \( d(2) = 4e^{4} - 0.18 \).- For the product \( p(t) \), \( p(2) = 0.72e^4 \).- For the quotient \( q(t) \), \( q(2) \approx 22.22e^4 \).Whether for checking work, understanding how a function behaves, or solving real-world problems, evaluating functions is foundational in both math and science. It provides specific insights and helps students see how theoretical calculations translate into concrete numbers. Next time you come across a function, remember that evaluating it at particular values can reveal much more than its general form might suggest!

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Most popular questions from this chapter

Lead Concentration Because of past use of leaded gasoline, the concentration of lead in soil can be associated with how close the soil is to a heavily traveled road. The table Lead Concentration in Soil Near Roads $$ \begin{array}{|c|c|} \hline \begin{array}{c} \text { Distance } \\ \text { (meters) } \end{array} & \begin{array}{c} \text { Lead } \\ \text { (ppm) } \end{array} \\ \hline 5 & 90 \\ \hline 10 & 60 \\ \hline 15 & 40 \\ \hline 20 & 32 \\ \hline \end{array} $$ shows average lead concentrations in parts per million of samples taken from different distances from roads. a. Find a log model for these data. b. An apartment complex has a dirt play area located 12 meters from a road. Calculate the lead concentration in the soil of the play area. c. Find an exponential model for the data. Compare this model to the log model found in part \(a\). Which of the two models better displays the end behavior suggested by the context?

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