If $$f(x) = f(-x)$$ for all $$x$$, then which of the following functions CANNOT be $$f(x)$$?

Incorrect.
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Try $$f(-x)$$.
>$$f(-x) = (-x)^2 + 1$$
Since $$(-x)^2 = (-x)(-x) = x^2$$, you can see that
>$$f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)$$.
So $$f(-x) = f(x)$$. **POE** this answer.

Incorrect.
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Try $$f(-x)$$.
>$$f(-x) = -2(-x)^2$$
Since $$(-x)^2 = (-x)(-x) = x^2$$, you can see that
>$$f(-x) = -2(-x)^2= -2x^2 = f(x)$$.
So $$f(-x) = f(x)$$. **POE** this answer.

Correct.
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**Plugging In** $$x=1$$,
>$$f(1)=(1−1)^2 = 0$$, and
>$$f(-1)=(1−(-1))^2 = 2^2 = 4$$.
You can see that $$f(x)\ne f(-x)$$ for $$x=1$$. So this cannot be $$f(x)$$, because the question states that $$f(x)=f(-x)$$ for **all** values of $$x$$.
**Alternatively**, you could show that $$f(-x)\ne f(x)$$ algebraically.
>$$f(-x) = (1-(-x))^2 = (1+x)^2$$
It should be clear that $$(1-x)^2 \ne (1+x)^2$$, meaning that $$f(x) \ne f(-x)$$. This must be the correct answer.

Incorrect.
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Try $$f(-x)$$.
>$$f(-x) = (1-(-x))(1-x)$$
>$$f(-x) = (1+x)(1-x) = f(x)$$
So $$f(-x) = f(x)$$. **POE** this answer.

Incorrect.
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Try $$f(-x)$$.
>$$\displaystyle f(-x) = \frac{7}{(-x)^2 + 3}$$
Since $$(-x)^2 = (-x)(-x) = x^2$$, you can see that
>$$\displaystyle f(-x) = \frac{7}{(-x)^2 + 3} = \frac{7}{x^2 + 3} = f(x)$$.
So $$f(-x) = f(x)$$. **POE** this answer.

$$f(x) = x^2+1$$

$$f(x) = -2x^2$$

$$f(x) = (1-x)^2$$

$$f(x) = (1-x)(1+x)$$

$$\displaystyle f(x) = \frac{7}{x^2+3}$$