Math / Algebra

QuestionSuppose that the functions $f$ and $g$ are defined as follows. $\begin{array}{l} f(x) = -3x + 5 \\ g(x) = \sqrt{2x - 3} \end{array}$
Find $f \cdot g$ and $f+g$ . Then, give their domains using interval notation.
$(f \cdot g)(x) = (-3x + 5)(\sqrt{2x - 3})$
Domain of $f \cdot g: \left[\frac{3}{2}, \infty\right)$
$(f+g)(x) =$ $\square$
Domain of $f+g$ : $\square$

Studdy Solution

STEP 1

What is this asking? We're given two functions, $f(x)$ and $g(x)$ , and we need to find the product of the functions ( $f \cdot g$ ) and the sum of the functions ( $f + g$ ), along with their domains. Watch out! Remember that the domain of the combined functions will be limited by the domains of the individual functions!

STEP 2

1. Find $f \cdot g$ .
2. Find the domain of $f \cdot g$ .
3. Find $f + g$ .
4. Find the domain of $f + g$ .

STEP 3

To find $f \cdot g$ , we multiply the expressions for $f(x)$ and $g(x)$ together.
It's as simple as that!
So, $(f \cdot g)(x) = f(x) \cdot g(x)$ .

STEP 4

Let's substitute the given functions: $(f \cdot g)(x) = (-3x + 5) \cdot (\sqrt{2x - 3})$ .

STEP 5

The domain of $f(x) = -3x + 5$ is all real numbers, since it's a linear function.
We can write this in interval notation as $(-\infty, \infty)$ .

STEP 6

The domain of $g(x) = \sqrt{2x - 3}$ is restricted because we can't take the square root of a negative number.
So, we need $2x - 3 \ge 0$ .

STEP 7

To solve for $x$ , we add 3 to both sides: $2x \ge 3$ .

STEP 8

Then, we divide both sides by 2: $x \ge \frac{3}{2}$ .
In interval notation, this is $[\frac{3}{2}, \infty)$ .

STEP 9

Since the domain of $f \cdot g$ is the intersection of the domains of $f$ and $g$ , the domain of $f \cdot g$ is $[\frac{3}{2}, \infty)$ .

STEP 10

To find $f + g$ , we add the expressions for $f(x)$ and $g(x)$ .
So, $(f + g)(x) = f(x) + g(x)$ .

STEP 11

Let's substitute the given functions: $(f + g)(x) = (-3x + 5) + (\sqrt{2x - 3})$ .

STEP 12

Similar to what we did before, the domain of $f + g$ is the intersection of the domains of $f$ and $g$ .

STEP 13

Since the domain of $f$ is all real numbers and the domain of $g$ is $x \ge \frac{3}{2}$ , the domain of $f + g$ is also $[\frac{3}{2}, \infty)$ .

STEP 14

$(f \cdot g)(x) = (-3x + 5)(\sqrt{2x - 3})$ with a domain of $[\frac{3}{2}, \infty)$ .
$(f + g)(x) = -3x + 5 + \sqrt{2x - 3}$ with a domain of $[\frac{3}{2}, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find $f \cdot g$ .
2. Find the domain of $f \cdot g$ .
3. Find $f + g$ .
4. Find the domain of $f + g$ .

STEP 3

To find $f \cdot g$ , we multiply the expressions for $f(x)$ and $g(x)$ together.
It's as simple as that!
So, $(f \cdot g)(x) = f(x) \cdot g(x)$ .

STEP 4

Let's substitute the given functions: $(f \cdot g)(x) = (-3x + 5) \cdot (\sqrt{2x - 3})$ .

STEP 5

The domain of $f(x) = -3x + 5$ is all real numbers, since it's a linear function.
We can write this in interval notation as $(-\infty, \infty)$ .

STEP 6

The domain of $g(x) = \sqrt{2x - 3}$ is restricted because we can't take the square root of a negative number.
So, we need $2x - 3 \ge 0$ .

STEP 7

To solve for $x$ , we add 3 to both sides: $2x \ge 3$ .

STEP 8

Then, we divide both sides by 2: $x \ge \frac{3}{2}$ .
In interval notation, this is $[\frac{3}{2}, \infty)$ .

STEP 9

Since the domain of $f \cdot g$ is the intersection of the domains of $f$ and $g$ , the domain of $f \cdot g$ is $[\frac{3}{2}, \infty)$ .

STEP 10

To find $f + g$ , we add the expressions for $f(x)$ and $g(x)$ .
So, $(f + g)(x) = f(x) + g(x)$ .

STEP 11

Let's substitute the given functions: $(f + g)(x) = (-3x + 5) + (\sqrt{2x - 3})$ .

STEP 12

Similar to what we did before, the domain of $f + g$ is the intersection of the domains of $f$ and $g$ .

STEP 13

Since the domain of $f$ is all real numbers and the domain of $g$ is $x \ge \frac{3}{2}$ , the domain of $f + g$ is also $[\frac{3}{2}, \infty)$ .

STEP 14

$(f \cdot g)(x) = (-3x + 5)(\sqrt{2x - 3})$ with a domain of $[\frac{3}{2}, \infty)$ .
$(f + g)(x) = -3x + 5 + \sqrt{2x - 3}$ with a domain of $[\frac{3}{2}, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find f⋅gf \cdot gf⋅g.2. Find the domain of f⋅gf \cdot gf⋅g.3. Find f+gf + gf+g.4. Find the domain of f+gf + gf+g.

STEP 3

To find f⋅gf \cdot gf⋅g, we **multiply** the expressions for f(x)f(x)f(x) and g(x)g(x)g(x) together.It's as simple as that!So, (f⋅g)(x)=f(x)⋅g(x)(f \cdot g)(x) = f(x) \cdot g(x)(f⋅g)(x)=f(x)⋅g(x).

STEP 4

Let's substitute the given functions: (f⋅g)(x)=(−3x+5)⋅(2x−3)(f \cdot g)(x) = (-3x + 5) \cdot (\sqrt{2x - 3})(f⋅g)(x)=(−3x+5)⋅(2x−3​).

STEP 5

The **domain** of f(x)=−3x+5f(x) = -3x + 5f(x)=−3x+5 is all real numbers, since it's a linear function.We can write this in interval notation as (−∞,∞)(-\infty, \infty)(−∞,∞).

STEP 6

The **domain** of g(x)=2x−3g(x) = \sqrt{2x - 3}g(x)=2x−3​ is restricted because we can't take the square root of a negative number.So, we need 2x−3≥02x - 3 \ge 02x−3≥0.

STEP 7

To solve for xxx, we **add** 3 to both sides: 2x≥32x \ge 32x≥3.

STEP 8

Then, we **divide** both sides by 2: x≥32x \ge \frac{3}{2}x≥23​.In interval notation, this is [32,∞)[\frac{3}{2}, \infty)[23​,∞).

STEP 9

Since the domain of f⋅gf \cdot gf⋅g is the intersection of the domains of fff and ggg, the domain of f⋅gf \cdot gf⋅g is [32,∞)[\frac{3}{2}, \infty)[23​,∞).

STEP 10

To find f+gf + gf+g, we **add** the expressions for f(x)f(x)f(x) and g(x)g(x)g(x).So, (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x)(f+g)(x)=f(x)+g(x).

STEP 11

Let's substitute the given functions: (f+g)(x)=(−3x+5)+(2x−3)(f + g)(x) = (-3x + 5) + (\sqrt{2x - 3})(f+g)(x)=(−3x+5)+(2x−3​).

STEP 12

Similar to what we did before, the domain of f+gf + gf+g is the intersection of the domains of fff and ggg.

STEP 13

Since the domain of fff is all real numbers and the domain of ggg is x≥32x \ge \frac{3}{2}x≥23​, the domain of f+gf + gf+g is also [32,∞)[\frac{3}{2}, \infty)[23​,∞).

STEP 14

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1. Find $f \cdot g$ .
2. Find the domain of $f \cdot g$ .
3. Find $f + g$ .
4. Find the domain of $f + g$ .

To find $f \cdot g$ , we multiply the expressions for $f(x)$ and $g(x)$ together.
It's as simple as that!
So, $(f \cdot g)(x) = f(x) \cdot g(x)$ .

Let's substitute the given functions: $(f \cdot g)(x) = (-3x + 5) \cdot (\sqrt{2x - 3})$ .

The domain of $f(x) = -3x + 5$ is all real numbers, since it's a linear function.
We can write this in interval notation as $(-\infty, \infty)$ .

The domain of $g(x) = \sqrt{2x - 3}$ is restricted because we can't take the square root of a negative number.
So, we need $2x - 3 \ge 0$ .

To solve for $x$ , we add 3 to both sides: $2x \ge 3$ .

Then, we divide both sides by 2: $x \ge \frac{3}{2}$ .
In interval notation, this is $[\frac{3}{2}, \infty)$ .

Since the domain of $f \cdot g$ is the intersection of the domains of $f$ and $g$ , the domain of $f \cdot g$ is $[\frac{3}{2}, \infty)$ .

To find $f + g$ , we add the expressions for $f(x)$ and $g(x)$ .
So, $(f + g)(x) = f(x) + g(x)$ .

Let's substitute the given functions: $(f + g)(x) = (-3x + 5) + (\sqrt{2x - 3})$ .

Similar to what we did before, the domain of $f + g$ is the intersection of the domains of $f$ and $g$ .

Since the domain of $f$ is all real numbers and the domain of $g$ is $x \ge \frac{3}{2}$ , the domain of $f + g$ is also $[\frac{3}{2}, \infty)$ .