+6251 \(f(x) = a x^6 - b x^4 + x - 1\\ \text{little trick here}\\ f_e(x) = a x^6 - b x^4 - 1 \text{ (this is the even part of }f(x))\\ f_e(x) = f_e(-x)\)
\(f(x) = x + f_e(x)\\ f(2) = 2 + f_e(2) = 5\\ f_e(2) = 3\\ f(-2) = -2 + f_e(-2) = \\ -2 + f_e(2) = \\ -2 + 3 = 1\)
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\(\text{if }f(x)=ax^6-bx^4+x-1\text{ and }f(2)=5\text{ what is }f(-2)?\)
The answer is actually 1.
\(\begin{align*} f(2)&=a(2^6)-b(2^4)+2-1 \\ 5&=a(64)-b(16)+1 \\ 4&=a(64)-b(16) \\ \end{align*}\)
Then,
\(\begin{align*} f(-2)&=a((-2)^6)-b((-2)^4)-2-1 \\ f(-2)&=a(64)-b(16)-3~~~(\leftarrow\text{We found what }64a-16b\text{ was in the previous equation.)} \\ f(-2)&=4-3 \\ f(-2)&=1 \end{align*}\)
