+23 .\(The/ function /f/ satisfies/ f(\sqrt{x + 1}) = \frac{1}{x} /for/ all/x \ge -1, x\neq 0. Find f(2).\)
2.\(Let/ f(x) = 3x^2 - 4x. Find/ the/ constant/ k/ such/ that/ f(x) = f(k - x) for/ all/ real /numbers/ x.\)
3.\(Find/ all/ complex /numbers/ z/ such /that/ z^2 = 2i. Write /your /solutions/ in/ a+bi /form/, separated/ by/ commas./ So,/ "1+2i, 3-i" /is/ an/ acceptable/ answer/ format,/ but/ "2i+1; -i+3"/ is/ not./ (Don't/ include/ quotes/ in/ your/ answer.)/ Note: /This/ problem /is /not /about/ functions.\)
4.
\(Let/ f/ be /a /function/ such/ that/ f(x+y) = x + f(y) /for/ any/ two/ real/ numbers/ x/ and/ y/. If f(0) = 2, then/ what/ is/ f(2012)?\)
+9479 1. The function \(f\) satisfies \(f(\sqrt{x+1})=\frac1x\) for all \(x\geq-1\,,\quad x\neq0\) . Find \(f(2)\) .
We want to find an x value such that...
\(\sqrt{x+1}=2\\~\\ x+1=4\\~\\ x=3\)
So......
\(f(\sqrt{x+1})=\frac1x\\~\\ f(\sqrt{3+1})=\frac13\\~\\ f(\sqrt{4})=\frac13\\~\\ f(2)=\frac13\)
+9479 1. The function \(f\) satisfies \(f(\sqrt{x+1})=\frac1x\) for all \(x\geq-1\,,\quad x\neq0\) . Find \(f(2)\) .
We want to find an x value such that...
\(\sqrt{x+1}=2\\~\\ x+1=4\\~\\ x=3\)
So......
\(f(\sqrt{x+1})=\frac1x\\~\\ f(\sqrt{3+1})=\frac13\\~\\ f(\sqrt{4})=\frac13\\~\\ f(2)=\frac13\)
+9479 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
I had to fix it!!
