Consider a function f:[0,2π]→R given by f(x)=sinx and g:[0,2π]→R given by g(x)=cosx. Show that f and g are one-one, but f+g is not one-one.
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Step-by-Step Solution
Step 1: Show f(x) is one-one
To show that a function is one-one, we assume that f(x1)=f(x2) for two values x1 and x2 in the domain. If this assumption implies that x1=x2, then the function is one-one. For f(x)=sinx on the interval [0,2π], if sinx1=sinx2, then x1=x2 because the sine function is strictly increasing in this interval.
Step 2: Show g(x) is one-one
Similarly, for g(x)=cosx on the interval [0,2π], if cosx1=cosx2, then x1=x2. This is because the cosine function is strictly decreasing in this interval, meaning each output value corresponds to a unique input value.
Step 3: Define (f+g)(x)
The sum of the two functions, (f+g)(x), is defined as the sum of their individual expressions. So, (f+g)(x)=sinx+cosx. We need to check if this new function is one-one on the given domain.
Step 4: Test if (f+g)(x) is one-one
To show that (f+g)(x) is not one-one, we need to find two distinct values x1 and x2 in the domain [0,2π] such that (f+g)(x1)=(f+g)(x2) but x1=x2. Let's evaluate the function at the endpoints of the domain. We find that (f+g)(0)=1 and (f+g)(2π)=1. Since 0=2π but their function values are equal, (f+g)(x) is not one-one.