Question

One-to-One Functions If \(f\) is a one-to-one function, provethat \(g(x)=-f(x)\) is also one-to-one.

Short answer

By following the definition of a one-to-one function, we understand that if \(f\) is a one-to-one function, then \(g(x) = -f(x)\) is also a one-to-one function.

Step-by-step solution

Understand the definition of one-to-one function

One-to-one, or injective, function is a function for which every element of the range of the function corresponds to exactly one element of the domain. So, if \(f\) is a one-to-one function, then whenever \(f(a) = f(b)\) for some \(a\) and \(b\) in the domain, it must be that \(a = b\).

Show that \(g(x)\) is a one-to-one function

We are required to prove that \(g(x) = -f(x)\) is also one-to-one. Suppose we have two inputs \(x_1\) and \(x_2\) where \(g(x_1) = g(x_2)\). This means that \(-f(x_1) = -f(x_2)\). We multiply both sides by -1 to get \(f(x_1) = f(x_2)\). Because we know that \(f\) is a one-to-one function, we know that if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\). Hence, we have proven that \(g(x) = -f(x)\) is also a one-to-one function.