The question is: is (root(x2+2))/x=f(x) even or odd?
I got that it was odd because f(-x)=-f(x) but on the graph it does not look like it is symmetric to the origin
what did i do wrong?
sqrt (x^2+2)
----------
x
The top will be the same for x positive or negative.
However the bottom will change sign but have the same magnitude
Therefore
Odd
check
If x = +5 then
sqrt (27)
---------
5
If x = -5 then
sqrt (27)
-------------
-5
But if you graph it. the graph is not symmetric to the origin. odd functions are always symm to the origin
Gee, it is for me, try -2 and +2
sqrt 6
---------
-2
and
sqrt 6
--------
+2
it is anti-symmetric as It should be
To determine whether the function f(x) = (sqrt(x^2 + 2))/x is even or odd, we need to analyze its symmetry.
A function is even if it satisfies the condition f(-x) = f(x) for all values of x in its domain. This means that if you replace x with -x in the function equation, the resulting expression should be equivalent to the original function.
On the other hand, a function is odd if it satisfies the condition f(-x) = -f(x) for all values of x in its domain. In this case, when you replace x with -x, the resulting expression should be the negative of the original function.
Let's now apply these conditions to the given function:
f(x) = (sqrt(x^2 + 2))/x
To determine whether it is even, we substitute -x for x and evaluate:
f(-x) = (sqrt((-x)^2 + 2))/(-x)
= (sqrt(x^2 + 2))/(-x)
Now let's compare this with the original function:
-f(x) = -((sqrt(x^2 + 2))/x)
= -((sqrt(x^2 + 2))/x)
By comparing f(-x) with -f(x), we can see that they are equal, not negative. Therefore, the function f(x) = (sqrt(x^2 + 2))/x is neither even nor odd.
It's worth noting that the appearance of the graph is not a reliable indicator of the function's symmetry properties. Symmetry can only be determined through algebraic calculations and comparisons.