Determine whether the functions are even, odd, or neither.
(a) f(x) = x²/√x²+ 1
(b) g(x) = x^3 - √ x- 7
Odd functions are symmetrical about the origin. Let f(x) be an odd function. f(-x) = -f(x). If its an odd function, it will satisfy this condition.
Even functions are symmetrical about that y axis. Let g(x) be an even function. g(-x) = g(x). If its an even function, it will satisfy this condition.
Neither is when it doesn't satisfy the conditions above.
I'll do (a). I'll assuming it's an even function.
g(-x) = g(x)
(-x)^2/sqrt((-x)^2+1) = x^2/sqrt((-x)^2+1)
x^2/sqrt(x^2+1) = x^2/sqrt(x^2+1)
Therefore, it is even.
You can always plug it in your calculator and check for symmetry.
To determine whether a function is even, odd, or neither, we need to analyze the function's symmetry with respect to the y-axis (even function) or the origin (odd function).
(a) For function f(x) = x²/√(x² + 1):
To check whether it is even, we need to replace x with -x and see if the function remains unchanged.
f(-x) = (-x)²/√((-x)² + 1)
= x²/√(x² + 1)
Comparing f(x) and f(-x), we can see that they are the same. This means that f(x) is an even function.
To check whether it is odd, we need to determine if replacing x with -x causes the function to change sign.
-f(-x) = -((-x)²/√((-x)² + 1))
= -x²/√(x² + 1)
Comparing f(x) and -f(-x), we can see that they differ in sign, indicating that f(x) is neither odd nor even.
Therefore, (a) f(x) = x²/√(x² + 1) neither even nor odd.
(b) For function g(x) = x³ - √(x - 7):
To check for evenness, replace x with -x and see if the function remains unchanged.
g(-x) = (-x)³ - √((-x) - 7)
= -x³ - √(-x - 7)
Comparing g(x) and g(-x), we can see that they are different. Therefore, g(x) is not an even function.
To check for oddness, replace x with -x and see if the function changes sign.
-g(-x) = -((-x)³ - √((-x) - 7))
= -(-x³ + √(x - 7))
= x³ - √(x - 7)
Comparing g(x) and -g(-x), we can see that they are the same. This indicates that g(x) is an odd function.
Therefore, (b) g(x) = x³ - √(x - 7) is an odd function.