Answer :
Certainly! Let's work through the problem step-by-step to understand the exponential function that decays at one-half the rate and evaluate it over the given interval [tex]\([-2, 0]\)[/tex].
### Step 1: Understand the Exponential Decay Function
Let's start by defining an exponential decay function [tex]\( f(x) \)[/tex]. A general exponential decay function can be written in the form:
[tex]\[ f(x) = e^{kx} \][/tex]
where [tex]\( k \)[/tex] is a constant.
### Step 2: Determine the Decay Rate
The problem states that the exponential function decays at one-half the rate. If the standard decay rate is 1, then a decay at one-half the rate would imply that [tex]\( k = -0.5 \)[/tex].
So, our function is:
[tex]\[ f(x) = e^{-0.5x} \][/tex]
### Step 3: Evaluate the Function at the Interval Bounds
We need to evaluate the function [tex]\( f(x) \)[/tex] at the lower bound [tex]\( x = -2 \)[/tex] and the upper bound [tex]\( x = 0 \)[/tex].
#### Lower Bound: [tex]\( x = -2 \)[/tex]
Substitute [tex]\( x = -2 \)[/tex] into the function:
[tex]\[ f(-2) = e^{-0.5(-2)} = e^1 = e \][/tex]
Based on the value of [tex]\( e \)[/tex]:
[tex]\[ f(-2) = 2.718281828459045 \][/tex]
#### Upper Bound: [tex]\( x = 0 \)[/tex]
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = e^{-0.5(0)} = e^0 = 1 \][/tex]
### Step 4: Conclusion
The value of the exponential function [tex]\( f(x) = e^{-0.5x} \)[/tex] evaluated at the lower bound [tex]\( x = -2 \)[/tex] is approximately 2.718281828459045, and at the upper bound [tex]\( x = 0 \)[/tex] is exactly 1.
Thus, the function values over the interval [tex]\([-2, 0]\)[/tex] are:
[tex]\[ f(-2) \approx 2.718281828459045 \][/tex]
[tex]\[ f(0) = 1 \][/tex]
### Step 1: Understand the Exponential Decay Function
Let's start by defining an exponential decay function [tex]\( f(x) \)[/tex]. A general exponential decay function can be written in the form:
[tex]\[ f(x) = e^{kx} \][/tex]
where [tex]\( k \)[/tex] is a constant.
### Step 2: Determine the Decay Rate
The problem states that the exponential function decays at one-half the rate. If the standard decay rate is 1, then a decay at one-half the rate would imply that [tex]\( k = -0.5 \)[/tex].
So, our function is:
[tex]\[ f(x) = e^{-0.5x} \][/tex]
### Step 3: Evaluate the Function at the Interval Bounds
We need to evaluate the function [tex]\( f(x) \)[/tex] at the lower bound [tex]\( x = -2 \)[/tex] and the upper bound [tex]\( x = 0 \)[/tex].
#### Lower Bound: [tex]\( x = -2 \)[/tex]
Substitute [tex]\( x = -2 \)[/tex] into the function:
[tex]\[ f(-2) = e^{-0.5(-2)} = e^1 = e \][/tex]
Based on the value of [tex]\( e \)[/tex]:
[tex]\[ f(-2) = 2.718281828459045 \][/tex]
#### Upper Bound: [tex]\( x = 0 \)[/tex]
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = e^{-0.5(0)} = e^0 = 1 \][/tex]
### Step 4: Conclusion
The value of the exponential function [tex]\( f(x) = e^{-0.5x} \)[/tex] evaluated at the lower bound [tex]\( x = -2 \)[/tex] is approximately 2.718281828459045, and at the upper bound [tex]\( x = 0 \)[/tex] is exactly 1.
Thus, the function values over the interval [tex]\([-2, 0]\)[/tex] are:
[tex]\[ f(-2) \approx 2.718281828459045 \][/tex]
[tex]\[ f(0) = 1 \][/tex]
