# What is the inverse function of f(x)= absx + 1?

Aug 16, 2015

No inverse as $f \left(x\right)$ is not a one-to-one function. However, it is possible to obtain a multivalued inverse.

Multivalued inverse:
${f}^{- 1} \left(x\right) = \pm \left(x - 1\right) , \text{ } x \ge 1$

#### Explanation:

Multivalued inverse:
$f \left(x\right) = | x | + 1$
$f \left(x\right) - 1 = | x |$
${f}^{- 1} \left(x\right) = \pm \left(x - 1\right) , \text{ } x \ge 1$

${f}^{- 1} \left(x\right)$ can also be found by taking y = x as the axis of symmetry.
Note that the range of f(x) becomes the domain of ${f}^{- 1} \left(x\right)$.

Aug 16, 2015

As a function from $\mathbb{R}$ to $\mathbb{R}$, $f \left(x\right)$ is not one-one.

As a result it has no well defined inverse function.

#### Explanation:

For example, $f \left(1\right) = f \left(- 1\right) = 2$, so ${f}^{- 1} \left(2\right)$ is not well defined - it could be $1$ or $- 1$.

If we restrict the domain of $f \left(x\right)$ to $\left[0 , \infty\right)$ or to $\left(- \infty , 0\right]$, then the resulting mapping is one-one onto $\left[1 , \infty\right)$ and there is a well defined inverse function with domain $\left[1 , \infty\right)$.

${f}_{\left[0 , \infty\right)} \left(x\right)$ has inverse ${f}_{\left[0 , \infty\right)}^{-} 1 \left(y\right) = y - 1$ where $y \in \left[1 , \infty\right)$

${f}_{\left(- \infty , 0\right]} \left(x\right)$ has inverse ${f}_{\left(- \infty , 0\right]}^{-} 1 \left(y\right) = 1 - y$ where $y \in \left[1 , \infty\right)$