Math / Algebra

QuestionLet $f(x)=4 x+3$ and $g(x)=2 x-7$ . Find $(f+g)(x),(f-g)(x),(f g)(x)$ , and $\left(\frac{f}{g}\right)(x)$ . Give the domain of each. $(f+g)(x)=$ $\square$ (Simplify your answer.)

Studdy Solution

STEP 1

What is this asking? We're given two functions, $f(x)$ and $g(x)$ , and we need to find their sum, difference, product, and quotient, along with the domain of each new function. Watch out! Be careful with the signs when subtracting and remember to consider any values that would make the denominator zero when dividing.

STEP 2

1. Find $(f+g)(x)$ and its domain.
2. Find $(f-g)(x)$ and its domain.
3. Find $(f \cdot g)(x)$ and its domain.
4. Find $\left(\frac{f}{g}\right)(x)$ and its domain.

STEP 3

We have $f(x) = 4x + 3$ and $g(x) = 2x - 7$ .

STEP 4

To find $(f+g)(x)$ , we simply add the expressions for $f(x)$ and $g(x)$ : $(f+g)(x) = f(x) + g(x) = (4x + 3) + (2x - 7)$

STEP 5

Combining like terms, we get: $(f+g)(x) = (4x + 2x) + (3 - 7) = 6x - 4$

STEP 6

Since $(f+g)(x) = 6x - 4$ is a linear function, its domain is all real numbers.
We can write this as $(-\infty, \infty)$ .

STEP 7

$(f-g)(x) = f(x) - g(x) = (4x + 3) - (2x - 7)$

STEP 8

Be extra careful with the signs! $(f-g)(x) = 4x + 3 - 2x + 7$

STEP 9

$(f-g)(x) = (4x - 2x) + (3 + 7) = 2x + 10$

STEP 10

The domain of $(f-g)(x) = 2x + 10$ , a linear function, is also all real numbers, $(-\infty, \infty)$ .

STEP 11

$(f \cdot g)(x) = f(x) \cdot g(x) = (4x + 3)(2x - 7)$

STEP 12

Using the FOIL method (First, Outer, Inner, Last): $(f \cdot g)(x) = 4x \cdot 2x + 4x \cdot (-7) + 3 \cdot 2x + 3 \cdot (-7)$ $(f \cdot g)(x) = 8x^2 - 28x + 6x - 21 = 8x^2 - 22x - 21$

STEP 13

Since $(f \cdot g)(x)$ is a quadratic function, its domain is all real numbers, $(-\infty, \infty)$ .

STEP 14

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{4x + 3}{2x - 7}$

STEP 15

The domain of $\left(\frac{f}{g}\right)(x)$ is all real numbers except for any value of $x$ that makes the denominator equal to zero.
We set the denominator equal to zero and solve for $x$ : $2x - 7 = 0$ $2x = 7$ $x = \frac{7}{2}$ So, the domain is all real numbers except $x = \frac{7}{2}$ .
We can write this as $\left(-\infty, \frac{7}{2}\right) \cup \left(\frac{7}{2}, \infty\right)$ .

STEP 16

$(f+g)(x) = 6x - 4$ , domain: $(-\infty, \infty)$ $(f-g)(x) = 2x + 10$ , domain: $(-\infty, \infty)$ $(f \cdot g)(x) = 8x^2 - 22x - 21$ , domain: $(-\infty, \infty)$ $\left(\frac{f}{g}\right)(x) = \frac{4x + 3}{2x - 7}$ , domain: $\left(-\infty, \frac{7}{2}\right) \cup \left(\frac{7}{2}, \infty\right)$

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QuestionLet $f(x)=4 x+3$ and $g(x)=2 x-7$ . Find $(f+g)(x),(f-g)(x),(f g)(x)$ , and $\left(\frac{f}{g}\right)(x)$ . Give the domain of each. $(f+g)(x)=$ $\square$ (Simplify your answer.)

Studdy Solution

STEP 1

STEP 2

1. Find $(f+g)(x)$ and its domain.
2. Find $(f-g)(x)$ and its domain.
3. Find $(f \cdot g)(x)$ and its domain.
4. Find $\left(\frac{f}{g}\right)(x)$ and its domain.

STEP 3

We have $f(x) = 4x + 3$ and $g(x) = 2x - 7$ .

STEP 4

To find $(f+g)(x)$ , we simply add the expressions for $f(x)$ and $g(x)$ : $(f+g)(x) = f(x) + g(x) = (4x + 3) + (2x - 7)$

STEP 5

Combining like terms, we get: $(f+g)(x) = (4x + 2x) + (3 - 7) = 6x - 4$

STEP 6

Since $(f+g)(x) = 6x - 4$ is a linear function, its domain is all real numbers.
We can write this as $(-\infty, \infty)$ .

STEP 7

$(f-g)(x) = f(x) - g(x) = (4x + 3) - (2x - 7)$

STEP 8

Be extra careful with the signs! $(f-g)(x) = 4x + 3 - 2x + 7$

STEP 9

$(f-g)(x) = (4x - 2x) + (3 + 7) = 2x + 10$

STEP 10

The domain of $(f-g)(x) = 2x + 10$ , a linear function, is also all real numbers, $(-\infty, \infty)$ .

STEP 11

$(f \cdot g)(x) = f(x) \cdot g(x) = (4x + 3)(2x - 7)$

STEP 12

Using the FOIL method (First, Outer, Inner, Last): $(f \cdot g)(x) = 4x \cdot 2x + 4x \cdot (-7) + 3 \cdot 2x + 3 \cdot (-7)$ $(f \cdot g)(x) = 8x^2 - 28x + 6x - 21 = 8x^2 - 22x - 21$

STEP 13

Since $(f \cdot g)(x)$ is a quadratic function, its domain is all real numbers, $(-\infty, \infty)$ .

STEP 14

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{4x + 3}{2x - 7}$

STEP 15

STEP 16

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Studdy Solution

STEP 1

STEP 2

1. Find (f+g)(x)(f+g)(x)(f+g)(x) and its domain.2. Find (f−g)(x)(f-g)(x)(f−g)(x) and its domain.3. Find (f⋅g)(x)(f \cdot g)(x)(f⋅g)(x) and its domain.4. Find (fg)(x)\left(\frac{f}{g}\right)(x)(gf​)(x) and its domain.

STEP 3

We have f(x)=4x+3f(x) = 4x + 3f(x)=4x+3 and g(x)=2x−7g(x) = 2x - 7g(x)=2x−7.

STEP 4

To find (f+g)(x)(f+g)(x)(f+g)(x), we simply add the expressions for f(x)f(x)f(x) and g(x)g(x)g(x): (f+g)(x)=f(x)+g(x)=(4x+3)+(2x−7)(f+g)(x) = f(x) + g(x) = (4x + 3) + (2x - 7)(f+g)(x)=f(x)+g(x)=(4x+3)+(2x−7)

STEP 5

Combining like terms, we get: (f+g)(x)=(4x+2x)+(3−7)=6x−4(f+g)(x) = (4x + 2x) + (3 - 7) = 6x - 4(f+g)(x)=(4x+2x)+(3−7)=6x−4

STEP 6

Since (f+g)(x)=6x−4(f+g)(x) = 6x - 4(f+g)(x)=6x−4 is a linear function, its domain is all real numbers.We can write this as (−∞,∞)(-\infty, \infty)(−∞,∞).

STEP 7

(f−g)(x)=f(x)−g(x)=(4x+3)−(2x−7)(f-g)(x) = f(x) - g(x) = (4x + 3) - (2x - 7)(f−g)(x)=f(x)−g(x)=(4x+3)−(2x−7)

STEP 8

Be extra careful with the signs! (f−g)(x)=4x+3−2x+7(f-g)(x) = 4x + 3 - 2x + 7(f−g)(x)=4x+3−2x+7

STEP 9

(f−g)(x)=(4x−2x)+(3+7)=2x+10(f-g)(x) = (4x - 2x) + (3 + 7) = 2x + 10(f−g)(x)=(4x−2x)+(3+7)=2x+10

STEP 10

The domain of (f−g)(x)=2x+10(f-g)(x) = 2x + 10(f−g)(x)=2x+10, a linear function, is also all real numbers, (−∞,∞)(-\infty, \infty)(−∞,∞).

STEP 11

(f⋅g)(x)=f(x)⋅g(x)=(4x+3)(2x−7)(f \cdot g)(x) = f(x) \cdot g(x) = (4x + 3)(2x - 7)(f⋅g)(x)=f(x)⋅g(x)=(4x+3)(2x−7)

STEP 12

STEP 13

Since (f⋅g)(x)(f \cdot g)(x)(f⋅g)(x) is a quadratic function, its domain is all real numbers, (−∞,∞)(-\infty, \infty)(−∞,∞).

STEP 14

(fg)(x)=f(x)g(x)=4x+32x−7\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{4x + 3}{2x - 7}(gf​)(x)=g(x)f(x)​=2x−74x+3​

STEP 15

STEP 16

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1. Find $(f+g)(x)$ and its domain.
2. Find $(f-g)(x)$ and its domain.
3. Find $(f \cdot g)(x)$ and its domain.
4. Find $\left(\frac{f}{g}\right)(x)$ and its domain.

We have $f(x) = 4x + 3$ and $g(x) = 2x - 7$ .

To find $(f+g)(x)$ , we simply add the expressions for $f(x)$ and $g(x)$ : $(f+g)(x) = f(x) + g(x) = (4x + 3) + (2x - 7)$

Combining like terms, we get: $(f+g)(x) = (4x + 2x) + (3 - 7) = 6x - 4$

Since $(f+g)(x) = 6x - 4$ is a linear function, its domain is all real numbers.
We can write this as $(-\infty, \infty)$ .

$(f-g)(x) = f(x) - g(x) = (4x + 3) - (2x - 7)$

Be extra careful with the signs! $(f-g)(x) = 4x + 3 - 2x + 7$

$(f-g)(x) = (4x - 2x) + (3 + 7) = 2x + 10$

The domain of $(f-g)(x) = 2x + 10$ , a linear function, is also all real numbers, $(-\infty, \infty)$ .

$(f \cdot g)(x) = f(x) \cdot g(x) = (4x + 3)(2x - 7)$

Since $(f \cdot g)(x)$ is a quadratic function, its domain is all real numbers, $(-\infty, \infty)$ .

$\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{4x + 3}{2x - 7}$