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Math  /  Algebra

QuestionFind the range. Write in interval notation. g(z)=4z+6−1g(z)=\sqrt{4 z+6}-1
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Studdy Solution

STEP 1

What is this asking? We need to find all possible output values of the function g(z)=4z+6−1g(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+64z + 6 has to be greater than or equal to zero.
Let's write that down: 4z+6≥04z + 6 \ge 0

STEP 4

Now, let's **isolate** zz.
First, we'll **subtract** 66 from both sides: 4z≥−64z \ge -6 Next, we'll **divide** both sides by 44: z≥−64z \ge \frac{-6}{4} Simplifying the fraction gives us: z≥−32z \ge -\frac{3}{2}

STEP 5

So, zz can be any number greater than or equal to −32-\frac{3}{2}.
In interval notation, that's [−32,∞)[-\frac{3}{2}, \infty).

STEP 6

The smallest value 4z+64z + 6 can be is **zero**, which happens when z=−32z = -\frac{3}{2} (the start of our domain).
If we plug that into g(z)g(z), we get: g(−32)=4⋅(−32)+6−1g\left(-\frac{3}{2}\right) = \sqrt{4 \cdot \left(-\frac{3}{2}\right) + 6} - 1 g(−32)=−6+6−1g\left(-\frac{3}{2}\right) = \sqrt{-6 + 6} - 1g(−32)=0−1g\left(-\frac{3}{2}\right) = \sqrt{0} - 1g(−32)=0−1g\left(-\frac{3}{2}\right) = 0 - 1g(−32)=−1g\left(-\frac{3}{2}\right) = -1So, the **minimum value** of g(z)g(z) is −1-1.

STEP 7

As zz gets bigger, 4z+64z + 6 also gets bigger.
This means the square root of 4z+64z + 6 gets bigger, and therefore g(z)g(z) gets bigger too!
Since there's no limit to how big zz can get (remember the domain!), there's no limit to how big g(z)g(z) can get either.

STEP 8

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

STEP 9

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

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