Math / Algebra

Question3. (4.1) Consider the following relation. $\begin{array}{l} -2 x-2 y= 3 x-y \\ +2 x+h 1 \\ +2 x+2 \end{array}$
Answer: $\quad f(x)=$ $\qquad$ $-5 x$
Step 2. Evaluate the function found in the previous step at $x=1$ . $\begin{array}{l} -5(1)-1 y \\ -5-1 y \end{array}$
Answer: $\quad f(1)=$ $\qquad$ Step 3. Determine the implied domain of the function found in the first step. Express your answer in interval notation.

Studdy Solution

STEP 1

What is this asking? We've got an equation, and we need to rewrite it so $y$ is a function of $x$ , then plug in $x = 1$ , and finally, figure out all the possible $x$ values that work for our function. Watch out! Don't forget to consider all possible $x$ values when finding the domain!

STEP 2

1. Rewrite as a function
2. Evaluate at x = 1
3. Find the domain

STEP 3

We want to get $y$ by itself on one side of the equation.
We start with $-2x - 2y = 3x - y$ .
Let's add $2x$ to both sides to move the $-2x$ from the left to the right: $-2x - 2y + 2x = 3x - y + 2x$ $-2y = 5x - y$ Now, let’s add $y$ to both sides to move the $-y$ from the right to the left: $-2y + y = 5x - y + y$ $-y = 5x$

STEP 4

Almost there!
We have $-y = 5x$ .
To get $y$ by itself, we multiply both sides by $-1$ : $(-1) \cdot (-y) = (-1) \cdot 5x$ $y = -5x$

STEP 5

So, our function is $f(x) = -5x$ !

STEP 6

Now, we plug in $x = 1$ into our function $f(x) = -5x$ : $f(1) = -5 \cdot 1$

STEP 7

$f(1) = -5$ So, $f(1)$ is $-5$ !

STEP 8

Our function is $f(x) = -5x$ .
Are there any $x$ values that would "break" this function?
Nope! We can multiply any real number by $-5$ .

STEP 9

Since any real number works for $x$ , the domain is all real numbers, which we write in interval notation as $(-\infty, \infty)$ .

STEP 10

$f(x) = -5x$ . $f(1) = -5$ .
The domain of $f(x)$ is $(-\infty, \infty)$ .

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Studdy Solution

STEP 1

What is this asking? We've got an equation, and we need to rewrite it so $y$ is a function of $x$ , then plug in $x = 1$ , and finally, figure out all the possible $x$ values that work for our function. Watch out! Don't forget to consider all possible $x$ values when finding the domain!

STEP 2

1. Rewrite as a function
2. Evaluate at x = 1
3. Find the domain

STEP 3

STEP 4

Almost there!
We have $-y = 5x$ .
To get $y$ by itself, we multiply both sides by $-1$ : $(-1) \cdot (-y) = (-1) \cdot 5x$ $y = -5x$

STEP 5

So, our function is $f(x) = -5x$ !

STEP 6

Now, we plug in $x = 1$ into our function $f(x) = -5x$ : $f(1) = -5 \cdot 1$

STEP 7

$f(1) = -5$ So, $f(1)$ is $-5$ !

STEP 8

Our function is $f(x) = -5x$ .
Are there any $x$ values that would "break" this function?
Nope! We can multiply any real number by $-5$ .

STEP 9

Since any real number works for $x$ , the domain is all real numbers, which we write in interval notation as $(-\infty, \infty)$ .

STEP 10

$f(x) = -5x$ . $f(1) = -5$ .
The domain of $f(x)$ is $(-\infty, \infty)$ .

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Studdy Solution

STEP 1

What is this asking? We've got an equation, and we need to rewrite it so yyy is a function of xxx, then plug in x=1x = 1x=1, and finally, figure out all the possible xxx values that work for our function. Watch out! Don't forget to consider *all* possible xxx values when finding the domain!

STEP 2

1. Rewrite as a function2. Evaluate at *x* = 13. Find the domain

STEP 3

STEP 4

Almost there!We have −y=5x-y = 5x−y=5x.To get yyy by itself, we multiply both sides by −1-1−1: (−1)⋅(−y)=(−1)⋅5x(-1) \cdot (-y) = (-1) \cdot 5x(−1)⋅(−y)=(−1)⋅5x y=−5xy = -5xy=−5x

STEP 5

So, our function is f(x)=−5xf(x) = -5xf(x)=−5x!

STEP 6

Now, we **plug in** x=1x = 1x=1 into our function f(x)=−5xf(x) = -5xf(x)=−5x: f(1)=−5⋅1f(1) = -5 \cdot 1f(1)=−5⋅1

STEP 7

f(1)=−5f(1) = -5f(1)=−5 So, f(1)f(1)f(1) is −5-5−5!

STEP 8

Our function is f(x)=−5xf(x) = -5xf(x)=−5x.Are there any xxx values that would "break" this function?Nope! We can multiply any real number by −5-5−5.

STEP 9

Since any real number works for xxx, the domain is all real numbers, which we write in interval notation as (−∞,∞)(-\infty, \infty)(−∞,∞).

STEP 10

f(x)=−5xf(x) = -5xf(x)=−5x. f(1)=−5f(1) = -5f(1)=−5.The domain of f(x)f(x)f(x) is (−∞,∞)(-\infty, \infty)(−∞,∞).

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What is this asking? We've got an equation, and we need to rewrite it so $y$ is a function of $x$ , then plug in $x = 1$ , and finally, figure out all the possible $x$ values that work for our function. Watch out! Don't forget to consider all possible $x$ values when finding the domain!

1. Rewrite as a function
2. Evaluate at x = 1
3. Find the domain

Almost there!
We have $-y = 5x$ .
To get $y$ by itself, we multiply both sides by $-1$ : $(-1) \cdot (-y) = (-1) \cdot 5x$ $y = -5x$

So, our function is $f(x) = -5x$ !

Now, we plug in $x = 1$ into our function $f(x) = -5x$ : $f(1) = -5 \cdot 1$

$f(1) = -5$ So, $f(1)$ is $-5$ !

Our function is $f(x) = -5x$ .
Are there any $x$ values that would "break" this function?
Nope! We can multiply any real number by $-5$ .

Since any real number works for $x$ , the domain is all real numbers, which we write in interval notation as $(-\infty, \infty)$ .

$f(x) = -5x$ . $f(1) = -5$ .
The domain of $f(x)$ is $(-\infty, \infty)$ .