Math Snap

STEP 1

What is this asking?
We need to find all possible output values of the function $g(z) = \sqrt{4z + 6} - 1$.
Basically, how high and how low can this thing go?!
Watch out!
Square roots can be tricky!
Remember, we can't take the square root of a negative number in the real number world.

STEP 2

1. Find the domain.
2. Find the range.

STEP 3

We know the stuff inside the square root can't be negative, so $4z + 6$ has to be greater than or equal to zero.
Let's write that down:
$$4z + 6 \ge 0$$

STEP 4

Now, let's isolate $z$.
First, we'll subtract $6$ from both sides:
$$4z \ge -6$$ Next, we'll divide both sides by $4$:
$$z \ge \frac{-6}{4}$$ Simplifying the fraction gives us:
$$z \ge -\frac{3}{2}$$

STEP 5

So, $z$ can be any number greater than or equal to $-\frac{3}{2}$.
In interval notation, that's $[-\frac{3}{2}, \infty)$.

STEP 6

The smallest value $4z + 6$ can be is zero, which happens when $z = -\frac{3}{2}$ (the start of our domain).
If we plug that into $g(z)$, we get:
$$g\left(-\frac{3}{2}\right) = \sqrt{4 \cdot \left(-\frac{3}{2}\right) + 6} - 1$$ $$g\left(-\frac{3}{2}\right) = \sqrt{-6 + 6} - 1$$$$g\left(-\frac{3}{2}\right) = \sqrt{0} - 1$$$$g\left(-\frac{3}{2}\right) = 0 - 1$$$$g\left(-\frac{3}{2}\right) = -1$$So, the minimum value of $g(z)$ is $-1$.

STEP 7

As $z$ gets bigger, $4z + 6$ also gets bigger.
This means the square root of $4z + 6$ gets bigger, and therefore $g(z)$ gets bigger too!
Since there's no limit to how big $z$ can get (remember the domain!), there's no limit to how big $g(z)$ can get either.

STEP 8

The smallest value $g(z)$ can be is $-1$, and it can go up to infinity!
In interval notation, the range is $[-1, \infty)$.

The range of $g(z)$ is $[-1, \infty)$.