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# Plugging In: DOZEN F for "Must Be" Questions

If $$f(x) = f(-x)$$ for all $$x$$, then which of the following functions CANNOT be $$f(x)$$?
Incorrect. [[snippet]] 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. [[snippet]] 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. [[snippet]] **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. [[snippet]] 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. [[snippet]] 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.
$$f(x) = x^2+1$$
$$f(x) = -2x^2$$
$$f(x) = (1-x)^2$$
$$f(x) = (1-x)(1+x)$$
$$\displaystyle f(x) = \frac{7}{x^2+3}$$