f(x)= x^2/√x^2 + 1
I have no clue.
You have defined the function f(x), but not asked a question. Your writing of the equation is unclear about whether x^2 +1 is all under the square root sign and in the denominator.
Oh, sorry.
I have to know whether it is even, odd, or neither.
Even because it satisfy f(x) = f(-x). In order words, symmetrical about the y axis.
If sqrt (x^2 +1) is the denominator, and if the positive square root is intended, then f(x) = f(-x), so the function is even.
No worries! Let's break down the given function step by step to understand it better.
The given function is:
f(x) = x^2 / √(x^2 + 1)
1. The function f(x) is defined as a fraction where the numerator is x^2 and the denominator is √(x^2 + 1).
2. In the denominator, we have the expression √(x^2 + 1). The square root symbol (√) indicates that we need to find the square root of the quantity (x^2 + 1). The quantity inside the square root is (x^2 + 1).
3. In the numerator, we have the expression x^2. This simply means that we need to square the value of x.
To evaluate this function for a particular value of x, you need to substitute that value into the function and perform the necessary calculations. For example, if you want to find f(2), you would substitute x = 2 into the function:
f(2) = (2^2) / √((2^2) + 1)
Now you can simplify the expression and calculate the result:
f(2) = 4 / √(4 + 1)
f(2) = 4 / √5
To get a numerical approximation for f(2), you would calculate the square root of 5 and divide 4 by that value.