Connect with a global community of experts on IDNLearn.com. Get comprehensive and trustworthy answers to all your questions from our knowledgeable community members.
Sagot :
To understand the function [tex]\( f(x) = e^x + 2 \)[/tex], we will go through its properties and provide detailed steps on how to evaluate it.
### Step-by-Step Explanation:
1. Function Definition:
The function [tex]\( f(x) \)[/tex] is defined as:
[tex]\[ f(x) = e^x + 2 \][/tex]
Here, [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828, and [tex]\( x \)[/tex] is the input variable.
2. Exponential Component:
The term [tex]\( e^x \)[/tex] represents the exponential function. Exponential functions grow rapidly and their rate of growth increases as the input [tex]\( x \)[/tex] increases.
3. Constant Addition:
The term [tex]\( +2 \)[/tex] is a constant that is added to [tex]\( e^x \)[/tex]. This constant shifts the entire exponential graph of [tex]\( e^x \)[/tex] upward by 2 units.
4. Function Evaluation:
To evaluate the function at a specific point, say [tex]\( x = a \)[/tex], substitute [tex]\( a \)[/tex] into the function:
[tex]\[ f(a) = e^a + 2 \][/tex]
This means you calculate the exponential part [tex]\( e^a \)[/tex] first and then add 2 to it.
### Examples:
- Example 1: Evaluate [tex]\( f(0) \)[/tex].
[tex]\[ f(0) = e^0 + 2 = 1 + 2 = 3 \][/tex]
Therefore, [tex]\( f(0) = 3 \)[/tex].
- Example 2: Evaluate [tex]\( f(1) \)[/tex].
[tex]\[ f(1) = e^1 + 2 \approx 2.71828 + 2 = 4.71828 \][/tex]
Therefore, [tex]\( f(1) \approx 4.71828 \)[/tex].
- Example 3: Evaluate [tex]\( f(-1) \)[/tex].
[tex]\[ f(-1) = e^{-1} + 2 \approx 0.36788 + 2 = 2.36788 \][/tex]
Therefore, [tex]\( f(-1) \approx 2.36788 \)[/tex].
### Graphical Representation:
The graph of [tex]\( f(x) = e^x + 2 \)[/tex] looks like the graph of [tex]\( e^x \)[/tex] but shifted upward by 2 units. It starts from just above 2 (when [tex]\( x \)[/tex] is very negative), crosses the y-axis at 3 (when [tex]\( x = 0 \)[/tex]), and rises exponentially as [tex]\( x \)[/tex] increases.
### Domain and Range:
- Domain: The domain of [tex]\( f(x) \)[/tex] is all real numbers, [tex]\( x \in (-\infty, \infty) \)[/tex]. This is because there are no restrictions on the values that [tex]\( x \)[/tex] can take for the exponential function [tex]\( e^x \)[/tex].
- Range: The range of [tex]\( f(x) \)[/tex] is [tex]\( (2, \infty) \)[/tex]. As [tex]\( e^x \)[/tex] always yields positive values and approaches 0 as [tex]\( x \)[/tex] goes to negative infinity, [tex]\( e^x + 2 \)[/tex] will always be greater than 2.
In summary, the function [tex]\( f(x) = e^x + 2 \)[/tex] is an exponential function shifted upwards by 2 units, with a domain of all real numbers and a range of values greater than 2.
### Step-by-Step Explanation:
1. Function Definition:
The function [tex]\( f(x) \)[/tex] is defined as:
[tex]\[ f(x) = e^x + 2 \][/tex]
Here, [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828, and [tex]\( x \)[/tex] is the input variable.
2. Exponential Component:
The term [tex]\( e^x \)[/tex] represents the exponential function. Exponential functions grow rapidly and their rate of growth increases as the input [tex]\( x \)[/tex] increases.
3. Constant Addition:
The term [tex]\( +2 \)[/tex] is a constant that is added to [tex]\( e^x \)[/tex]. This constant shifts the entire exponential graph of [tex]\( e^x \)[/tex] upward by 2 units.
4. Function Evaluation:
To evaluate the function at a specific point, say [tex]\( x = a \)[/tex], substitute [tex]\( a \)[/tex] into the function:
[tex]\[ f(a) = e^a + 2 \][/tex]
This means you calculate the exponential part [tex]\( e^a \)[/tex] first and then add 2 to it.
### Examples:
- Example 1: Evaluate [tex]\( f(0) \)[/tex].
[tex]\[ f(0) = e^0 + 2 = 1 + 2 = 3 \][/tex]
Therefore, [tex]\( f(0) = 3 \)[/tex].
- Example 2: Evaluate [tex]\( f(1) \)[/tex].
[tex]\[ f(1) = e^1 + 2 \approx 2.71828 + 2 = 4.71828 \][/tex]
Therefore, [tex]\( f(1) \approx 4.71828 \)[/tex].
- Example 3: Evaluate [tex]\( f(-1) \)[/tex].
[tex]\[ f(-1) = e^{-1} + 2 \approx 0.36788 + 2 = 2.36788 \][/tex]
Therefore, [tex]\( f(-1) \approx 2.36788 \)[/tex].
### Graphical Representation:
The graph of [tex]\( f(x) = e^x + 2 \)[/tex] looks like the graph of [tex]\( e^x \)[/tex] but shifted upward by 2 units. It starts from just above 2 (when [tex]\( x \)[/tex] is very negative), crosses the y-axis at 3 (when [tex]\( x = 0 \)[/tex]), and rises exponentially as [tex]\( x \)[/tex] increases.
### Domain and Range:
- Domain: The domain of [tex]\( f(x) \)[/tex] is all real numbers, [tex]\( x \in (-\infty, \infty) \)[/tex]. This is because there are no restrictions on the values that [tex]\( x \)[/tex] can take for the exponential function [tex]\( e^x \)[/tex].
- Range: The range of [tex]\( f(x) \)[/tex] is [tex]\( (2, \infty) \)[/tex]. As [tex]\( e^x \)[/tex] always yields positive values and approaches 0 as [tex]\( x \)[/tex] goes to negative infinity, [tex]\( e^x + 2 \)[/tex] will always be greater than 2.
In summary, the function [tex]\( f(x) = e^x + 2 \)[/tex] is an exponential function shifted upwards by 2 units, with a domain of all real numbers and a range of values greater than 2.
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
