Math / Algebra

QuestionUse the tables to evaluate the following expressions, if possible. \begin{tabular}{|c|c|c|c|} \hline $x$ & -2 & 2 & 3 \\ \hline $f(x)$ & -2 & 2 & 1 \\ \hline \end{tabular} (a) $(f+g)(3)$ \begin{tabular}{|c|c|c|c|} \hline $\mathbf{x}$ & -2 & 2 & 3 \\ \hline $\mathbf{g}(\mathbf{x})$ & 0 & 4 & 3 \\ \hline \end{tabular} (c) $(g f)(-2)$ (b) $(f-g)(2)$ (d) $\left(\frac{f}{g}\right)(-2)$ (a) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. $(f+g)(3)=$ $\square$ B. The answer is undefined.

Studdy Solution

STEP 1

What is this asking? We're looking up function values in tables and then doing some arithmetic with them!
Specifically, we need to find $(f+g)(3)$ , $(f-g)(2)$ , $(gf)(-2)$ , and $\left(\frac{f}{g}\right)(-2)$ . Watch out! Make sure you're looking up the right x-values in the right tables!
Also, remember what those function operation symbols mean: $(f+g)(x)$ means $f(x) + g(x)$ , $(f-g)(x)$ means $f(x) - g(x)$ , $(gf)(x)$ means $f(x) \cdot g(x)$ , and $\left(\frac{f}{g}\right)(x)$ means $\frac{f(x)}{g(x)}$ .

STEP 2

1. Calculate $(f+g)(3)$
2. Calculate $(f-g)(2)$
3. Calculate $(gf)(-2)$
4. Calculate $\left(\frac{f}{g}\right)(-2)$

STEP 3

Alright, let's start with $(f+g)(3)$ .
Remember, this just means we need to find $f(3)$ and $g(3)$ , and then add them together!

STEP 4

Looking at the tables, when $x = \mathbf{3}$ , we see that $f(\mathbf{3}) = \mathbf{1}$ and $g(\mathbf{3}) = \mathbf{3}$ .

STEP 5

So, $(f+g)(3) = f(3) + g(3) = \mathbf{1} + \mathbf{3} = \mathbf{4}$ .
Awesome!

STEP 6

Next up: $(f-g)(2)$ .
This means we find $f(2)$ and $g(2)$ , and then subtract $g(2)$ from $f(2)$ .

STEP 7

From the tables, when $x = \mathbf{2}$ , we have $f(\mathbf{2}) = \mathbf{2}$ and $g(\mathbf{2}) = \mathbf{4}$ .

STEP 8

Therefore, $(f-g)(2) = f(2) - g(2) = \mathbf{2} - \mathbf{4} = \mathbf{-2}$ .
Fantastic!

STEP 9

Now for $(gf)(-2)$ .
This means we need to multiply $f(-2)$ and $g(-2)$ .

STEP 10

The tables tell us that when $x = \mathbf{-2}$ , $f(\mathbf{-2}) = \mathbf{-2}$ and $g(\mathbf{-2}) = \mathbf{0}$ .

STEP 11

So, $(gf)(-2) = f(-2) \cdot g(-2) = \mathbf{-2} \cdot \mathbf{0} = \mathbf{0}$ .
Excellent!

STEP 12

Finally, let's tackle $\left(\frac{f}{g}\right)(-2)$ .
This means we need to divide $f(-2)$ by $g(-2)$ .

STEP 13

We already know from the previous step that $f(\mathbf{-2}) = \mathbf{-2}$ and $g(\mathbf{-2}) = \mathbf{0}$ .

STEP 14

So, $\left(\frac{f}{g}\right)(-2) = \frac{f(-2)}{g(-2)} = \frac{\mathbf{-2}}{\mathbf{0}}$ .
Uh oh!
We can't divide by zero.

STEP 15

Since division by zero is undefined, $\left(\frac{f}{g}\right)(-2)$ is undefined.

STEP 16

(a) $(f+g)(3) = 4$ (b) $(f-g)(2) = -2$ (c) $(gf)(-2) = 0$ (d) $\left(\frac{f}{g}\right)(-2)$ is undefined.

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Studdy Solution

STEP 1

STEP 2

1. Calculate $(f+g)(3)$
2. Calculate $(f-g)(2)$
3. Calculate $(gf)(-2)$
4. Calculate $\left(\frac{f}{g}\right)(-2)$

STEP 3

Alright, let's start with $(f+g)(3)$ .
Remember, this just means we need to find $f(3)$ and $g(3)$ , and then add them together!

STEP 4

Looking at the tables, when $x = \mathbf{3}$ , we see that $f(\mathbf{3}) = \mathbf{1}$ and $g(\mathbf{3}) = \mathbf{3}$ .

STEP 5

So, $(f+g)(3) = f(3) + g(3) = \mathbf{1} + \mathbf{3} = \mathbf{4}$ .
Awesome!

STEP 6

Next up: $(f-g)(2)$ .
This means we find $f(2)$ and $g(2)$ , and then subtract $g(2)$ from $f(2)$ .

STEP 7

From the tables, when $x = \mathbf{2}$ , we have $f(\mathbf{2}) = \mathbf{2}$ and $g(\mathbf{2}) = \mathbf{4}$ .

STEP 8

Therefore, $(f-g)(2) = f(2) - g(2) = \mathbf{2} - \mathbf{4} = \mathbf{-2}$ .
Fantastic!

STEP 9

Now for $(gf)(-2)$ .
This means we need to multiply $f(-2)$ and $g(-2)$ .

STEP 10

The tables tell us that when $x = \mathbf{-2}$ , $f(\mathbf{-2}) = \mathbf{-2}$ and $g(\mathbf{-2}) = \mathbf{0}$ .

STEP 11

So, $(gf)(-2) = f(-2) \cdot g(-2) = \mathbf{-2} \cdot \mathbf{0} = \mathbf{0}$ .
Excellent!

STEP 12

Finally, let's tackle $\left(\frac{f}{g}\right)(-2)$ .
This means we need to divide $f(-2)$ by $g(-2)$ .

STEP 13

We already know from the previous step that $f(\mathbf{-2}) = \mathbf{-2}$ and $g(\mathbf{-2}) = \mathbf{0}$ .

STEP 14

So, $\left(\frac{f}{g}\right)(-2) = \frac{f(-2)}{g(-2)} = \frac{\mathbf{-2}}{\mathbf{0}}$ .
Uh oh!
We can't divide by zero.

STEP 15

Since division by zero is undefined, $\left(\frac{f}{g}\right)(-2)$ is undefined.

STEP 16

(a) $(f+g)(3) = 4$ (b) $(f-g)(2) = -2$ (c) $(gf)(-2) = 0$ (d) $\left(\frac{f}{g}\right)(-2)$ is undefined.

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Studdy Solution

STEP 1

STEP 2

1. Calculate (f+g)(3)(f+g)(3)(f+g)(3)2. Calculate (f−g)(2)(f-g)(2)(f−g)(2)3. Calculate (gf)(−2)(gf)(-2)(gf)(−2)4. Calculate (fg)(−2) \left(\frac{f}{g}\right)(-2) (gf​)(−2)

STEP 3

Alright, let's **start** with (f+g)(3)(f+g)(3)(f+g)(3).Remember, this just means we need to find f(3)f(3)f(3) and g(3)g(3)g(3), and then **add** them together!

STEP 4

Looking at the tables, when x=3x = \mathbf{3}x=3, we see that f(3)=1f(\mathbf{3}) = \mathbf{1}f(3)=1 and g(3)=3g(\mathbf{3}) = \mathbf{3}g(3)=3.

STEP 5

So, (f+g)(3)=f(3)+g(3)=1+3=4(f+g)(3) = f(3) + g(3) = \mathbf{1} + \mathbf{3} = \mathbf{4}(f+g)(3)=f(3)+g(3)=1+3=4.Awesome!

STEP 6

Next up: (f−g)(2)(f-g)(2)(f−g)(2).This means we find f(2)f(2)f(2) and g(2)g(2)g(2), and then **subtract** g(2)g(2)g(2) from f(2)f(2)f(2).

STEP 7

From the tables, when x=2x = \mathbf{2}x=2, we have f(2)=2f(\mathbf{2}) = \mathbf{2}f(2)=2 and g(2)=4g(\mathbf{2}) = \mathbf{4}g(2)=4.

STEP 8

Therefore, (f−g)(2)=f(2)−g(2)=2−4=−2(f-g)(2) = f(2) - g(2) = \mathbf{2} - \mathbf{4} = \mathbf{-2}(f−g)(2)=f(2)−g(2)=2−4=−2.Fantastic!

STEP 9

Now for (gf)(−2)(gf)(-2)(gf)(−2).This means we need to **multiply** f(−2)f(-2)f(−2) and g(−2)g(-2)g(−2).

STEP 10

The tables tell us that when x=−2x = \mathbf{-2}x=−2, f(−2)=−2f(\mathbf{-2}) = \mathbf{-2}f(−2)=−2 and g(−2)=0g(\mathbf{-2}) = \mathbf{0}g(−2)=0.

STEP 11

So, (gf)(−2)=f(−2)⋅g(−2)=−2⋅0=0(gf)(-2) = f(-2) \cdot g(-2) = \mathbf{-2} \cdot \mathbf{0} = \mathbf{0}(gf)(−2)=f(−2)⋅g(−2)=−2⋅0=0.Excellent!

STEP 12

Finally, let's tackle (fg)(−2) \left(\frac{f}{g}\right)(-2) (gf​)(−2).This means we need to **divide** f(−2)f(-2)f(−2) by g(−2)g(-2)g(−2).

STEP 13

We already know from the previous step that f(−2)=−2f(\mathbf{-2}) = \mathbf{-2}f(−2)=−2 and g(−2)=0g(\mathbf{-2}) = \mathbf{0}g(−2)=0.

STEP 14

So, (fg)(−2)=f(−2)g(−2)=−20 \left(\frac{f}{g}\right)(-2) = \frac{f(-2)}{g(-2)} = \frac{\mathbf{-2}}{\mathbf{0}} (gf​)(−2)=g(−2)f(−2)​=0−2​.Uh oh!We can't divide by zero.

STEP 15

Since division by zero is undefined, (fg)(−2) \left(\frac{f}{g}\right)(-2) (gf​)(−2) is **undefined**.

STEP 16

(a) (f+g)(3)=4(f+g)(3) = 4(f+g)(3)=4 (b) (f−g)(2)=−2(f-g)(2) = -2(f−g)(2)=−2 (c) (gf)(−2)=0(gf)(-2) = 0(gf)(−2)=0 (d) (fg)(−2) \left(\frac{f}{g}\right)(-2) (gf​)(−2) is undefined.

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1. Calculate $(f+g)(3)$
2. Calculate $(f-g)(2)$
3. Calculate $(gf)(-2)$
4. Calculate $\left(\frac{f}{g}\right)(-2)$

Alright, let's start with $(f+g)(3)$ .
Remember, this just means we need to find $f(3)$ and $g(3)$ , and then add them together!

Looking at the tables, when $x = \mathbf{3}$ , we see that $f(\mathbf{3}) = \mathbf{1}$ and $g(\mathbf{3}) = \mathbf{3}$ .

So, $(f+g)(3) = f(3) + g(3) = \mathbf{1} + \mathbf{3} = \mathbf{4}$ .
Awesome!

Next up: $(f-g)(2)$ .
This means we find $f(2)$ and $g(2)$ , and then subtract $g(2)$ from $f(2)$ .

From the tables, when $x = \mathbf{2}$ , we have $f(\mathbf{2}) = \mathbf{2}$ and $g(\mathbf{2}) = \mathbf{4}$ .

Therefore, $(f-g)(2) = f(2) - g(2) = \mathbf{2} - \mathbf{4} = \mathbf{-2}$ .
Fantastic!

Now for $(gf)(-2)$ .
This means we need to multiply $f(-2)$ and $g(-2)$ .

The tables tell us that when $x = \mathbf{-2}$ , $f(\mathbf{-2}) = \mathbf{-2}$ and $g(\mathbf{-2}) = \mathbf{0}$ .

So, $(gf)(-2) = f(-2) \cdot g(-2) = \mathbf{-2} \cdot \mathbf{0} = \mathbf{0}$ .
Excellent!

Finally, let's tackle $\left(\frac{f}{g}\right)(-2)$ .
This means we need to divide $f(-2)$ by $g(-2)$ .

We already know from the previous step that $f(\mathbf{-2}) = \mathbf{-2}$ and $g(\mathbf{-2}) = \mathbf{0}$ .

So, $\left(\frac{f}{g}\right)(-2) = \frac{f(-2)}{g(-2)} = \frac{\mathbf{-2}}{\mathbf{0}}$ .
Uh oh!
We can't divide by zero.

Since division by zero is undefined, $\left(\frac{f}{g}\right)(-2)$ is undefined.

(a) $(f+g)(3) = 4$ (b) $(f-g)(2) = -2$ (c) $(gf)(-2) = 0$ (d) $\left(\frac{f}{g}\right)(-2)$ is undefined.