Let \(f(x) = x^n + k\), where \(n\) is a natural number and \(k\) is a positive real constant such that \(f(x) + f\left(\dfrac{1}{x}\right) = f(x)\,f\left(\dfrac{1}{x}\right)\) and \(f(2) = 9\). What is \(f(-1)\) equal to?
\(0\)
Substituting \(f(x)=x^n+k\) into \(f(x)+f(1/x)=f(x)f(1/x)\) and comparing coefficients gives \(k=1\), so \(f(x)=x^n+1\). Using \(f(2)=9\) gives \(2^n+1=9\), i.e. \(2^n=8\), so \(n=3\), giving \(f(x)=x^3+1\). Hence \(f(-1) = -1+1 = 0\).
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