Question

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Determine if the function is increasing, decreasing, even, odd, and/or invertible on its natural domain: f(x)=x^2-2x+3
State all the things that applies to it. For ex. in it's both increasing and decreasing then say both (don't leave out anything)

Answer 1

Answer:None

For $f(x) = x^2 - 2x + 3$, we have that $f(0) = 3$, $f(1) = 2$, and $f(2) = 3$, so $f(0) > f(1)$ and $f(1) < f(2)$. Hence, the function $f(x)$ is neither increasing nor decreasing. We have that

\[f(-x) = (-x)^2 - 2(-x) + 3 = x^2 + 2x + 3,\]

which is equal to neither $f(x)$ nor $-f(x)$ (which we can determine by plugging in, say, $x = 1$), so $f(x)$ is neither even nor odd.

From $f(0) = 3$ and $f(2) = 3$, we see that there are two different values of $x$ for which $f(x) = 3$. Therefore, $f(x)$ is not invertible

Answer 2

For the function $f(x) = x^2 - 2x + 3,$ f(x) is neither even nor odd.  f(x) is not invertible.

What is a function?

The function is a type of relation, or rule, that maps one input to a specific single output.

For the function $f(x) = x^2 - 2x + 3,$

we have that f(0) = 3,

f(1) = 2,

f(2) = 3,

So, f(0) > f(1) and f(1) < f(2).

Hence, the function f(x) is neither increasing nor decreasing.

We have that

$f(-x) = (-x)^2 - 2(-x) + 3 \\\\f(-x) = x^2 + 2x + 3,\]$

which is equal to neither f(x) nor -f(x)

So, f(x) is neither even nor odd.

From f(0) = 3

f(2) = 3,

we can see that there are two different values of x for which f(x) = 3.

Therefore, f(x) is not invertible.

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