Question

Show that \(f\) is one-to-one on the indicated interval and therefore has an inverse function on that interval. $$ f(x)=|x+2| \quad[-2, \infty) $$

Short answer

Yes, the function \(f(x)=|x+2|\) is one-to-one on the interval \([-2, \infty)\). Therefore, it has an inverse function on that interval, which we can denote as \(f^{-1}(x) = x-2\), and is defined on \([0, \infty)\).

Step-by-step solution

Understanding the Function

Firstly, let's understand the function \(f(x)=|x+2|\). This is a modulus function, which simply takes the absolute value of \(x+2\). The modulus of any number is always positive, implying that this function's output cannot be negative. On passing -2 to the function \(f(x)\), we get \(f(-2) = |-2+2| = 0\). Therefore, our function starts from 0 at \(x=-2\). Now, for all \(x > -2\), \(f(x)\) will be \(x+2\), as \(x+2\) is positive when \(x > -2\). So, we can rewrite \(f(x)\) for \([-2, \infty)\) as \(f(x) = x+2\).

Proving the Function is One-to-One

To prove the function is one-to-one, what we need to do is take two arbitrary points \(x1\) and \(x2\), where \(x1 != x2\). If the function \(f(x)\) gives different outputs for \(x1\) and \(x2\), then function \(f(x)\) is one-to-one. So, let's calculate \(f(x1)\) and \(f(x2)\): \(f(x1) = x1 + 2\), \(f(x2) = x2 + 2\). If \(x1 != x2\), then \(f(x1) != f(x2)\). Hence, function \(f(x)\) is one-to-one on the interval \([-2, \infty)\).

Determining the Inverse Function

As \(f(x)\) is one-to-one on the interval \([-2, \infty)\), it must have an inverse function. The way to find the inverse is to replace \(f(x)\) with \(y\), so \(y = x+2\). Then, swap \(x\) and \(y\), getting \(x = y+2\). Finally, solve for \(y\) to get the inverse function: \(y = x-2\). Therefore, the inverse function is \(f^{-1}(x) = x-2\) which is defined on \(f(x)\)'s range, i.e., \([0, \infty)\).