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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.

 Jun 9, 2022
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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, \(k=-\dfrac{3}{2}\)

 Jun 9, 2022

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