# What are invertible functions? plz explain.

Mar 26, 2018

See explanation...

#### Explanation:

There are at least two things implied by the assertion:

"$f \left(x\right)$ is invertible function $\forall x \in \left[1 , 5\right]$"

• $f \left(x\right)$ is a function with domain including all of the real interval $\left[1 , 5\right]$

• For any ${x}_{1} , {x}_{2} \in \left[1 , 5\right]$ we have $f \left({x}_{1}\right) = f \left({x}_{2}\right) \implies {x}_{1} = {x}_{2}$. In other words, $f \left(x\right)$ is one to one.

Considered as a function on the domain $\left[1 , 5\right]$, there is an inverse function $g \left(x\right)$ on the range of $f \left(x\right)$, such that:

$g \left(f \left(x\right)\right) = x \text{ }$ for all $x \in \left[1 , 5\right]$

$f \left(g \left(x\right)\right) = x \text{ }$ for all $x \in f \left(\text{["1, 5"]}\right)$

We are not told anything about the behaviour of $f \left(x\right)$ outside of this interval. For example, it could be undefined or not one to one.

For the purposes of the rest of the question we do not care.

Given:

$g \left(3\right) = 1$ and $g \left(6\right) = 5$

we can deduce:

$f \left(1\right) = 3$ and $f \left(5\right) = 6$

If the question is correct - which I am not sure - then we can find which multiple choice option is correct by using a linear function for $f \left(x\right)$ satisfying the conditions.

Let:

$f \left(x\right) = \frac{3}{4} \left(x - 1\right) + 3 = \frac{3}{4} x + \frac{9}{4}$

$g \left(x\right) = \frac{4}{3} \left(x - 3\right) + 1 = \frac{4}{3} x - 3$

Integrate each one the given intervals and add to find the answer.

I think the question is flawed in at least a couple of ways:

• The integration limits for $g \left(x\right)$ should be $\textcolor{red}{3}$ and $6$, not $5$ and $6$.

• Some extra conditions on the functions are required to make them integrable. For example, we could specify that $f \left(x\right)$ is continuous.

With the question corrected, the answer I would expect would be the sum of the areas of three rectangles:

• Vertices $\left(1 , 0\right) , \left(5 , 0\right) , \left(5 , 3\right) , \left(1 , 3\right)$ with area $4 \cdot 3 = 12$

• Vertices $\left(1 , 3\right) , \left(5 , 3\right) , \left(5 , 6\right) , \left(1 , 6\right)$ with area $4 \cdot 3 = 12$

• Vertices $\left(0 , 3\right) , \left(1 , 3\right) , \left(1 , 6\right) , \left(0 , 6\right)$ with area $1 \cdot 3 = 3$

So, total $12 + 12 + 3 = 27$, i.e. option 2)