18) Given the function $$ f(x)=x^{4}-x^{2}+3 $$, which of the following is true? $$ \begin{array}{ll} ext { A) }-\mathrm{f}(-x)=\mathrm{f}(x) & ext { C) } \mathrm{f}(x)=\mathrm{f}^{2}(x) \ ext { B) }-\mathrm{f}(x)=\mathrm{f}(x) & ext { D) } \mathrm{f}(-x)=\mathrm{f}(x)\end{array} $$

Question

18) Given the function $$ f(x)=x^{4}-x^{2}+3 $$, which of the following is true? $$ \begin{array}{ll}	ext { A) }-\mathrm{f}(-x)=\mathrm{f}(x) & 	ext { C) } \mathrm{f}(x)=\mathrm{f}^{2}(x) \ 	ext { B) }-\mathrm{f}(x)=\mathrm{f}(x) & 	ext { D) } \mathrm{f}(-x)=\mathrm{f}(x)\end{array} $$

UpStudy AI · Accepted Answer

Let's analyze each option for the function $$ f(x) = x^4 - x^2 + 3 $$:

**Option A: $$ -f(-x) = f(x) $$**  
First, compute $$ f(-x) $$:
$$ f(-x) = (-x)^4 - (-x)^2 + 3 = x^4 - x^2 + 3 = f(x) $$
So,
$$ -f(-x) = -f(x) $$
This implies $$ -f(x) = f(x) $$, which would only be true if $$ f(x) = 0 $$. However:
$$ f(0) = 0^4 - 0^2 + 3 = 3 \neq 0 $$
Thus, **Option A is false**.

**Option B: $$ -f(x) = f(x) $$**  
This suggests $$ f(x) = 0 $$ for all $$ x $$, which is not the case as:
$$ f(0) = 3 \neq 0 $$
Thus, **Option B is false**.

**Option C: $$ f(x) = f^2(x) $$**  
Assuming $$ f^2(x) $$ means $$ [f(x)]^2 $$:
$$ [f(x)]^2 = (x^4 - x^2 + 3)^2 $$
This is a quartic expression, which is generally not equal to the original quartic $$ f(x) $$ unless $$ f(x) = 0 $$ or $$ f(x) = 1 $$. However:
$$ f(0) = 3 \neq 1 $$
Thus, **Option C is false**.

**Option D: $$ f(-x) = f(x) $$**  
As calculated in Option A:
$$ f(-x) = f(x) $$
This shows that the function is even.

**Conclusion:** Only **Option D** is true.

**Answer:**  
Only statement D is true. $$ \mathrm{f}(-x) = \mathrm{f}(x) $$
Only statement D is true: $$ f(-x) = f(x) $$.

Given the function , which of the
following is true?

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