In Exercises 47–62, say whether the function is even, odd, or neither. Give reasons for your answer.
f(x) = x² + x
Verified step by step guidance
1
To determine if a function is even, odd, or neither, we need to analyze the function's symmetry properties. A function f(x) is even if f(-x) = f(x) for all x in the domain, and it is odd if f(-x) = -f(x) for all x in the domain.
Start by substituting -x into the function f(x) = x² + x to find f(-x). This gives us f(-x) = (-x)² + (-x).
Simplify the expression for f(-x). Since (-x)² = x², we have f(-x) = x² - x.
Compare f(-x) = x² - x with the original function f(x) = x² + x. Notice that f(-x) is not equal to f(x), so the function is not even.
Next, check if f(-x) = -f(x). Calculate -f(x) = -(x² + x) = -x² - x. Since f(-x) = x² - x is not equal to -f(x) = -x² - x, the function is not odd. Therefore, the function is neither even nor odd.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above
Video duration:
2m
Play a video:
0 Comments
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Even Functions
A function is classified as even if it satisfies the condition f(-x) = f(x) for all x in its domain. This means that the graph of the function is symmetric with respect to the y-axis. A common example of an even function is f(x) = x², where substituting -x yields the same output as substituting x.
A function is considered odd if it meets the condition f(-x) = -f(x) for all x in its domain. This indicates that the graph of the function is symmetric with respect to the origin. An example of an odd function is f(x) = x³, where substituting -x results in the negative of the output for x.
A function is classified as neither even nor odd if it does not satisfy the conditions for either classification. This means that f(-x) does not equal f(x) and also does not equal -f(x). An example is f(x) = x² + x, as it does not exhibit symmetry about the y-axis or the origin.