1. ## Functions

Let the function f : [3,1) -> R be defined by f(z) = (z −3)^2 +5, and
g : R -> R be defined by g(x) = 7 e^x.
(a) Find ran f.
(b) Find g(ran f).
(c) Let B := ran f and C := g(ran f). Write explicitly the expression for (g)^−1(h) for h E C,
and the expression for (f)^−1(x) for x E B.
(d) Write explicitly the expression for (g o f)^−1(h) for h E C.

2. Originally Posted by luckyc1423
Let the function f : [3,1) -> R be defined by f(z) = (z −3)^2 +5, and
g : R -> R be defined by g(x) = 7 e^x.
(a) Find ran f.
What is $\displaystyle [3,1)=\{\}$?
I assume $\displaystyle [1,3)$.

The range are all $\displaystyle y\in \mathbb{R}$ such that $\displaystyle f(z)=y$ for some $\displaystyle z\in [1,3)$.
Thus,
$\displaystyle f(z)=y$
$\displaystyle (z-3)^2+5=y$
$\displaystyle (z-3)^2=y-5$
To have a solution we need that, $\displaystyle y-5\geq 0$.
Thus,
$\displaystyle z=3\pm \sqrt{y-5}$
We need that,
$\displaystyle 1\leq 3\pm \sqrt{y-5} <3$
$\displaystyle -2\leq \pm \sqrt{y-5}<0$
We see that $\displaystyle +$ cannot work.
Thus,
$\displaystyle -2\leq -\sqrt{y-5}<0$ this tells us that $\displaystyle y\not = 5$.
It is equivalnet to squaring both sides,
$\displaystyle 4\geq y-5<0$
$\displaystyle 9\geq y < 5$
Note it satisfies the condition we have before $\displaystyle y-5\geq 0$.
(b) Find g(ran f).
This is defined as,
$\displaystyle g: (5,9]\to \mathbb{R}$
$\displaystyle g(x)=7e^x$