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10 questions from this quiz
Multiplying arguments
f(xy)f(\frac{x}{y})f(yx)
f(x)f(y)f(x)f(y)f(x)f(y)
Multiplying exponential values
f(x)=2xf(x) = 2xf(x)=2x
The constant term prevents it
f(x)+f(y)=(x+y)f(xy)f(x) + f(y) = (x + y)f(xy)f(x)+f(y)=(x+y)f(xy)
f(x)=1xf(x) = \frac{1}{x}f(x)=x1 is undefined at x=0x=0x=0
−1-1−1
One counterexample
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