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

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. \( \mathrm{f}(-x) = \mathrm{f}(x) \)