Let f(x)=3x^2 - 4x - x^2 + 7x + 8. Find the constant k such that f(x)=f(k-x) for all real numbers x.
Hi!
So first, simplify \(f(x)\) to get:
\(f(x)=2x^2+3x+8\)
Then, find \(f(k-x)\) as follows:
\(f(k-x)=2(k-x)^2+3(k-x)+8\)
\(=2(k^2-2kx+x^2)+3k-3x+8\)
\(=2x^2-4kx+2k^2+3k-3x+8\)
\(=2x^2+(-4k-3)x+(3k+8+2k^2)\)
Thus, \(f(k-x)=2x^2+(-4k-3)x+(2k^2+3k+8)\)
Then, set:
\(f(x)=f(k-x) \)
\(\iff 2x^2+3x+8=2x^2+(-4k-3)x+(2k^2+3k+8)\)
\(\iff 3x+8=(-4k-3)x+(2k^2+3k+8) \)
Equate coefficients of \(x:\)
\(3=-4k-3 \implies k=-\frac{3}{2}\)
Therefore, \(k=-\dfrac{3}{2}\)