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Math  /  Algebra

Question6. If f(x)=x−3f(x)=\sqrt{x-3} and g(x)=x+2g(x)=\sqrt{x+2} find (f−g)(x)(f-g)(x) and state its domain.

Studdy Solution

STEP 1

What is this asking? We're finding a new function by subtracting two square root functions, and then figuring out where that new function is defined! Watch out! Remember that the domain of a square root function is restricted to non-negative inputs.

STEP 2

1. Define the functions
2. Subtract the functions
3. Find the domain

STEP 3

We're given two functions, f(x)f(x) and g(x)g(x), both involving square roots.
Let's write them down clearly: f(x)=x−3f(x) = \sqrt{x-3} g(x)=x+2g(x) = \sqrt{x+2}It's **super important** to remember what a square root means.
We're looking for a number that, when multiplied by itself, gives us the value inside the square root.

STEP 4

We need to find (f−g)(x)(f-g)(x), which just means we **subtract** g(x)g(x) from f(x)f(x).
It's as simple as that! (f−g)(x)=f(x)−g(x)(f-g)(x) = f(x) - g(x) Let's substitute the expressions we have for f(x)f(x) and g(x)g(x): (f−g)(x)=x−3−x+2(f-g)(x) = \sqrt{x-3} - \sqrt{x+2} Boom! That's our new function!

STEP 5

Now, the **domain** is the set of all xx values where our new function makes sense.
Since we have square roots, we need to make sure what's *inside* each square root isn't negative.

STEP 6

For f(x)f(x), we need x−3≥0x-3 \ge 0.
Adding 3 to both sides gives us x≥3x \ge 3.

STEP 7

For g(x)g(x), we need x+2≥0x+2 \ge 0.
Subtracting 2 from both sides gives us x≥−2x \ge -2.

STEP 8

Since our new function (f−g)(x)(f-g)(x) involves *both* f(x)f(x) and g(x)g(x), we need *both* of those conditions to be true. xx has to be greater than or equal to 3 *and* greater than or equal to -2.
Since 3 is already bigger than -2, we only need x≥3x \ge 3.
So our domain is all xx values greater than or equal to **3**.
We can write this as [3,∞)[3, \infty).

STEP 9

The new function is (f−g)(x)=x−3−x+2(f-g)(x) = \sqrt{x-3} - \sqrt{x+2} and its domain is [3,∞)[3, \infty).

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