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Minchanka [31]
3 years ago
10

Is the function even, odd, or neither?

Mathematics
2 answers:
Step2247 [10]3 years ago
3 0

Answer:

A function with a graph that is symmetric about the origin is called an odd function. Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f(x)=2x f ( x ) = 2 x is neither even nor odd.

Step-by-step explanation:

enot [183]3 years ago
3 0

You may be asked to "determine algebraically" whether a function is even or odd. To do this, you take the function and plug –x in for x, and then simplify. If you end up with the exact same function that you started with (that is, if f (–x) = f (x), so all of the signs are the same), then the function is even. If you end up with the exact opposite of what you started with (that is, if f (–x) = –f (x), so all of the signs are switched), then the function is odd.

In all other cases, the function is "neither even nor odd".

Let's see what this looks like in action:

Determine algebraically whether f (x) = –3x2 + 4 is even, odd, or neither.

If I graph this, I will see that this is "symmetric about the y-axis"; in other words, whatever the graph is doing on one side of the y-axis is mirrored on the other side:

graph of y = –3x^2 + 4

This mirroring about the y-axis is a hallmark of even functions.

Also, I note that the exponents on all of the terms are even — the exponent on the constant term being zero: 4x0 = 4 × 1 = 4. These are helpful clues that strongly suggest to me that I've got an even function here.

But the question asks me to make the determination algebraically, which means that I need to do the algebra.

So I'll plug –x in for x, and simplify:

f (–x) = –3(–x)2 + 4

= –3(x2) + 4

= –3x2 + 4

I can see, by comparing the original function with my final result above, that I've got a match, which means that:

f (x) is even

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