If \(f\bigl(x^{2}\bigr)=x^{4}-6 x^{2}+5,\) which of the following must be true?

- \(f\bigl(x^{2}\bigr)=f\bigl(1+x^{2}\bigr)\)
- \(f\bigl(x^{2}\bigr)=f\bigl(2-x^{2}\bigr)\)
- \(f\bigl(x^{2}\bigr)=f\bigl(3-x^{2}\bigr)\)
- \(f\bigl(x^{2}\bigr)=f\bigl(5-x^{2}\bigr)\)
- \(f\bigl(x^{2}\bigr)=f\bigl(6-x^{2}\bigr)\)

The problem checks your understanding of plugging one expression \(g(x)\) into another \(f(x)\).

To substitute \(g(x) = x -1\) for \(x\) in \(f(x)= 2x^2 +4x +7\), I replace for clarity the dummy input variable by \(z\) and put protective parentheses around each of its occurrences. \[
\begin{align}
f(z) &= 2(z)^2 +4(z) +7 \\
f(x-1) &= 2(x-1)^2 +4(x-1) +7\,.
\end{align}
\] Expanding and simplifying yields \(f\bigl(g(x)\bigr) = 2x^2 +5\).

Observe that the relation \(z = x -1\) can be uniquely solved for \(x\), \[z = x -1 \Leftrightarrow x = z +1\,,\] the replacement \(g(x)\) is invertible, and the original function \(f(z)\) can be recovered by substituting \(x = z +1\): \[
f(z) = 2(z +1)^2 +5 = 2z^2 +4z +7
\]

The replacement \(g(x) = x^2\) isn’t invertible for all real \(x\). Only for \(x \ge 0\) we have \(z =x^2 \Leftrightarrow x = \sqrt{z}\). Therefore, we can’t recover \(f(z)\) completely. Indeed, even \[
f_1(z) = \begin{cases}
1 +4\cos z &\text{, if $z < 0$} \\
z^2 -6z +5 &\text{, if $z \ge 0$}
\end{cases}
\] yields \(f_1(x^2) = x^4 -6x^2 +5\) for all real \(x\). Only non-negative inputs to \(f\) are important!

GMAT problems don’t expect you to deal with this kind of intricacies, and it should be safe to take \(f(z) = z^2 -6z +5\) even for \(z < 0\). To recover \(f(z)\), I substituted \(z\) for \(x^2\) in \(f(x^2) = x^2\cdot x^2 -6\cdot x^2 +5\).

A statement like \(f\bigl(x^{2}\bigr)=f\bigl(2-x^{2}\bigr)\) has to be translated carefully, \[
f(x^2) \rightsquigarrow f(z) \rightsquigarrow f(2-x^2)\,,
\] and means substituting \(2-x^2\) for every occurrence of \(x^2\) in \(f(x^2) = x^2\cdot x^2 -6\cdot x^2 +5\): \[
\overbrace{x^4-6x^2 +5}^{f(x^2)} = \overbrace{(2 -x^2)^2-6(2-x^2) +5}^{f(2-x^2)}
\] We don’t have to solve here an equation. The equation is supposed to be true **for all** input values \(x\), and we only have to check if that’s true. While the test input \(x = 1\) does solve the equation, the input \(x = 0\) already shows that \(f(x^2) = f(2-x^2)\) is generally false, and we can dismiss answer choice (B).

Similarly, the test input \(x = 0\) doesn’t solve any of the equations \[
\begin{aligned}
x^4-6x^2 +5 &= (1 +x^2)^2-6(1+x^2) +5 \\
x^4-6x^2 +5 &= (3 -x^2)^2-6(3-x^2) +5 \\
x^4-6x^2 +5 &= (5 -x^2)^2-6(5-x^2) +5\,,
\end{aligned}
\] and we can reject answer choices (A), (C), and (D). The only option left is (E). Indeed, expanding \(f(6 -x^2)\) yields \[
(6 -x^2)^2-6(6 -x^2) +5 = 36 -12x^2 +x^4 -36 +6x^2 +5\,,
\] which is equivalent to \(x^4 -6x^2 +5\).

If the answer choices of a GMAT problem contain a variable \(x\) (function, equation, inequality), the answer choices are supposed to be **true for all values of \(x\)**. It’s usually cheap and fast to discard wrong answers by plugging in test values.