Question 1027846
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Let f(x) = 3x^2 - 4x. Find the constant k such that  f(x) = f(k - x) for all real numbers x.
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The condition requires that this identity would hold for all real numbers x:

{{{3x^2 - 4x}}} = {{{3*(k-x)^2 - 4(k-x)}}}.

Simplify it step by step

{{{3x^2 - 4x}}} = {{{3k^2 - 6kx + 3x^2 - 4k + 4x}}},

{{{3k^2 - 6kx - 4k + 8x}}} = {{{0}}}.

It would be identically zero if

{{{3k^2}}} = {{{4k}}}    (1)   and  
{{{6kx}}} = {{{8x}}}     (2)

are hold simultaneously.

Fortunately,  k = {{{4/3}}}  satisfies both  (1)  and  (2).

<U>Answer</U>.  k = {{{4/3}}}.
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