.The/function/f/satisfies/f(√x+1)=1x/for/all/x≥−1,x≠0.Findf(2).
2.Let/f(x)=3x2−4x.Find/the/constant/k/such/that/f(x)=f(k−x)for/all/real/numbers/x.
3.Find/all/complex/numbers/z/such/that/z2=2i.Write/your/solutions/in/a+bi/form/,separated/by/commas./So,/"1+2i,3−i"/is/an/acceptable/answer/format,/but/"2i+1;−i+3"/is/not./(Don′t/include/quotes/in/your/answer.)/Note:/This/problem/is/not/about/functions.
4.
Let/f/be/a/function/such/that/f(x+y)=x+f(y)/for/any/two/real/numbers/x/and/y/.Iff(0)=2,then/what/is/f(2012)?
2. Let f(x) = 3x2 - 4x . Find the constant k such that f(x) = f(k - x) for all real numbers x .
f(x) = f(k - x)
And f(x) = 3x2 - 4x
3x2 - 4x = f(k - x)
And f(k - x) = 3(k - x)2 - 4(k - x)
3x2 - 4x = 3(k - x)2 - 4(k - x)
3x2 - 3(k - x)2 - 4x + 4(k - x) = 0
3[x2 - (k - x)2] - 4[x - (k - x)] = 0
3[x + (k - x)][x - (k - x)] - 4[x - (k - x)] = 0
3[x + k - x][x - k + x] - 4[x - k + x] = 0
3[ k ][ 2x - k ] - 4[ 2x - k ] = 0
( 2x - k )( 3k - 4 ) = 0
2x - k = 0 or 3k - 4 = 0
k = 2x k = 4/3
The constant value that works is k = 4/3