Math / Algebra

Question12. If $f(x)$ and $g(x)$ are even functions, then what type of function is $y=f(x)-g(x)$ ? a. odd b. even c. neither d. cannot be determined for sure
13. To solve the inequality $f(x)>g(x)$ , a student could graph the combined function $y=f(x)-g(x)$ and identify the portions of the graph that are below the $x$ -axis. a) True b) false
14. If $f(x)$ and $g(x)$ are both functions that are defined for all $x \in \mathbb{R}$ , then $f(g(x))=g(f(x))$ . a) True b) false
15. If $f(x)$ is a function that is defined for all $x \in \mathbb{R}$ , then $f\left(f^{-1}(x)\right)=x$ . a) True b) false
Part B - Thinking and Investigation Full marks will be given only for if all steps are shown[TI - 15 marks]

Studdy Solution

STEP 1

What is this asking? These questions are testing our knowledge of function properties like even and odd functions, function composition, and inverse functions, plus how to solve inequalities graphically. Watch out! Don't mix up even/odd functions or confuse function composition with multiplication!
Also, remember that inverse functions don't always exist for all x.

STEP 2

1. Even/Odd Functions
2. Inequality Solutions
3. Function Composition
4. Inverse Functions

STEP 3

Let's break down what it means for $f(x)$ and $g(x)$ to be even functions.
Remember, a function is even if $f(-x) = f(x)$ for all $x$ .
Think of it like a mirror image across the y-axis!

STEP 4

Now, let's look at $y = f(x) - g(x)$ .
We want to see if this new function is even or odd.
So, we'll substitute $-x$ for $x$ : $y(-x) = f(-x) - g(-x)$ .

STEP 5

Since $f(x)$ and $g(x)$ are even, we know $f(-x) = f(x)$ and $g(-x) = g(x)$ .
So, we can replace those in our equation: $y(-x) = f(x) - g(x)$ .

STEP 6

Look closely! $y(-x) = f(x) - g(x)$ is the same as our original equation $y = f(x) - g(x)$ , which means $y(-x) = y(x)$ .
This is the definition of an even function!

STEP 7

The statement says that to solve $f(x) > g(x)$ , we can graph $y = f(x) - g(x)$ and look for where the graph is below the x-axis.
Is this true?

STEP 8

Let's think about it.
If $f(x) > g(x)$ , then subtracting $g(x)$ from both sides gives us $f(x) - g(x) > 0$ .

STEP 9

The combined function is $y = f(x) - g(x)$ .
We want to find where $f(x) - g(x) > 0$ , which means we're looking for where $y > 0$ .
This corresponds to the portions of the graph above the x-axis, not below!

STEP 10

The statement claims that $f(g(x)) = g(f(x))$ for all functions defined for all real numbers.
Let's test this with a simple example.

STEP 11

Let $f(x) = x + 1$ and $g(x) = 2x$ .
Then $f(g(x)) = f(2x) = 2x + 1$ .

STEP 12

Now let's find $g(f(x))$ : $g(f(x)) = g(x+1) = 2(x+1) = 2x + 2$ .

STEP 13

We see that $2x + 1$ is not equal to $2x + 2$ , so $f(g(x))$ is not equal to $g(f(x))$ in general.

STEP 14

The statement says that if $f(x)$ is defined for all $x$ , then $f(f^{-1}(x)) = x$ .
This is the definition of an inverse function, but there's a catch!

STEP 15

The inverse function, $f^{-1}(x)$ , must also be defined for all $x$ for this to be true.
If $f(x)$ doesn't have an inverse defined for all $x$ , then the statement isn't necessarily true.

STEP 16

12. b. even
13. b) false
14. b) false
15. b) false

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Studdy Solution

STEP 1

STEP 2

1. Even/Odd Functions
2. Inequality Solutions
3. Function Composition
4. Inverse Functions

STEP 3

Let's break down what it means for $f(x)$ and $g(x)$ to be even functions.
Remember, a function is even if $f(-x) = f(x)$ for all $x$ .
Think of it like a mirror image across the y-axis!

STEP 4

Now, let's look at $y = f(x) - g(x)$ .
We want to see if this new function is even or odd.
So, we'll substitute $-x$ for $x$ : $y(-x) = f(-x) - g(-x)$ .

STEP 5

Since $f(x)$ and $g(x)$ are even, we know $f(-x) = f(x)$ and $g(-x) = g(x)$ .
So, we can replace those in our equation: $y(-x) = f(x) - g(x)$ .

STEP 6

Look closely! $y(-x) = f(x) - g(x)$ is the same as our original equation $y = f(x) - g(x)$ , which means $y(-x) = y(x)$ .
This is the definition of an even function!

STEP 7

The statement says that to solve $f(x) > g(x)$ , we can graph $y = f(x) - g(x)$ and look for where the graph is below the x-axis.
Is this true?

STEP 8

Let's think about it.
If $f(x) > g(x)$ , then subtracting $g(x)$ from both sides gives us $f(x) - g(x) > 0$ .

STEP 9

The combined function is $y = f(x) - g(x)$ .
We want to find where $f(x) - g(x) > 0$ , which means we're looking for where $y > 0$ .
This corresponds to the portions of the graph above the x-axis, not below!

STEP 10

The statement claims that $f(g(x)) = g(f(x))$ for all functions defined for all real numbers.
Let's test this with a simple example.

STEP 11

Let $f(x) = x + 1$ and $g(x) = 2x$ .
Then $f(g(x)) = f(2x) = 2x + 1$ .

STEP 12

Now let's find $g(f(x))$ : $g(f(x)) = g(x+1) = 2(x+1) = 2x + 2$ .

STEP 13

We see that $2x + 1$ is not equal to $2x + 2$ , so $f(g(x))$ is not equal to $g(f(x))$ in general.

STEP 14

The statement says that if $f(x)$ is defined for all $x$ , then $f(f^{-1}(x)) = x$ .
This is the definition of an inverse function, but there's a catch!

STEP 15

The inverse function, $f^{-1}(x)$ , must also be defined for all $x$ for this to be true.
If $f(x)$ doesn't have an inverse defined for all $x$ , then the statement isn't necessarily true.

STEP 16

12. b. even
13. b) false
14. b) false
15. b) false

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Studdy Solution

STEP 1

STEP 2

1. Even/Odd Functions2. Inequality Solutions3. Function Composition4. Inverse Functions

STEP 3

Let's **break down** what it means for f(x)f(x)f(x) and g(x)g(x)g(x) to be even functions.Remember, a function is even if f(−x)=f(x)f(-x) = f(x)f(−x)=f(x) for all xxx.Think of it like a mirror image across the y-axis!

STEP 4

Now, let's look at y=f(x)−g(x)y = f(x) - g(x)y=f(x)−g(x).We want to see if this new function is even or odd.So, we'll **substitute** −x-x−x for xxx: y(−x)=f(−x)−g(−x)y(-x) = f(-x) - g(-x)y(−x)=f(−x)−g(−x).

STEP 5

Since f(x)f(x)f(x) and g(x)g(x)g(x) are even, we know f(−x)=f(x)f(-x) = f(x)f(−x)=f(x) and g(−x)=g(x)g(-x) = g(x)g(−x)=g(x).So, we can **replace** those in our equation: y(−x)=f(x)−g(x)y(-x) = f(x) - g(x)y(−x)=f(x)−g(x).

STEP 6

Look closely! y(−x)=f(x)−g(x)y(-x) = f(x) - g(x)y(−x)=f(x)−g(x) is the same as our original equation y=f(x)−g(x)y = f(x) - g(x)y=f(x)−g(x), which means y(−x)=y(x)y(-x) = y(x)y(−x)=y(x).This is the **definition** of an even function!

STEP 7

The statement says that to solve f(x)>g(x)f(x) > g(x)f(x)>g(x), we can graph y=f(x)−g(x)y = f(x) - g(x)y=f(x)−g(x) and look for where the graph is below the x-axis.Is this true?

STEP 8

Let's **think** about it.If f(x)>g(x)f(x) > g(x)f(x)>g(x), then subtracting g(x)g(x)g(x) from both sides gives us f(x)−g(x)>0f(x) - g(x) > 0f(x)−g(x)>0.

STEP 9

The combined function is y=f(x)−g(x)y = f(x) - g(x)y=f(x)−g(x).We want to find where f(x)−g(x)>0f(x) - g(x) > 0f(x)−g(x)>0, which means we're looking for where y>0y > 0y>0.This corresponds to the portions of the graph *above* the x-axis, not below!

STEP 10

The statement claims that f(g(x))=g(f(x))f(g(x)) = g(f(x))f(g(x))=g(f(x)) for all functions defined for all real numbers.Let's **test** this with a simple example.

STEP 11

Let f(x)=x+1f(x) = x + 1f(x)=x+1 and g(x)=2xg(x) = 2xg(x)=2x.Then f(g(x))=f(2x)=2x+1f(g(x)) = f(2x) = 2x + 1f(g(x))=f(2x)=2x+1.

STEP 12

Now let's find g(f(x))g(f(x))g(f(x)): g(f(x))=g(x+1)=2(x+1)=2x+2g(f(x)) = g(x+1) = 2(x+1) = 2x + 2g(f(x))=g(x+1)=2(x+1)=2x+2.

STEP 13

We see that 2x+12x + 12x+1 is not equal to 2x+22x + 22x+2, so f(g(x))f(g(x))f(g(x)) is not equal to g(f(x))g(f(x))g(f(x)) in general.

STEP 14

The statement says that if f(x)f(x)f(x) is defined for all xxx, then f(f−1(x))=xf(f^{-1}(x)) = xf(f−1(x))=x.This is the **definition** of an inverse function, but there's a catch!

STEP 15

The inverse function, f−1(x)f^{-1}(x)f−1(x), must also be defined for all xxx for this to be true.If f(x)f(x)f(x) doesn't have an inverse defined for all xxx, then the statement isn't necessarily true.

STEP 16

12. b. even13. b) false14. b) false15. b) false

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1. Even/Odd Functions
2. Inequality Solutions
3. Function Composition
4. Inverse Functions

Let's break down what it means for $f(x)$ and $g(x)$ to be even functions.
Remember, a function is even if $f(-x) = f(x)$ for all $x$ .
Think of it like a mirror image across the y-axis!

Now, let's look at $y = f(x) - g(x)$ .
We want to see if this new function is even or odd.
So, we'll substitute $-x$ for $x$ : $y(-x) = f(-x) - g(-x)$ .

Since $f(x)$ and $g(x)$ are even, we know $f(-x) = f(x)$ and $g(-x) = g(x)$ .
So, we can replace those in our equation: $y(-x) = f(x) - g(x)$ .

Look closely! $y(-x) = f(x) - g(x)$ is the same as our original equation $y = f(x) - g(x)$ , which means $y(-x) = y(x)$ .
This is the definition of an even function!

The statement says that to solve $f(x) > g(x)$ , we can graph $y = f(x) - g(x)$ and look for where the graph is below the x-axis.
Is this true?

Let's think about it.
If $f(x) > g(x)$ , then subtracting $g(x)$ from both sides gives us $f(x) - g(x) > 0$ .

The combined function is $y = f(x) - g(x)$ .
We want to find where $f(x) - g(x) > 0$ , which means we're looking for where $y > 0$ .
This corresponds to the portions of the graph above the x-axis, not below!

The statement claims that $f(g(x)) = g(f(x))$ for all functions defined for all real numbers.
Let's test this with a simple example.

Let $f(x) = x + 1$ and $g(x) = 2x$ .
Then $f(g(x)) = f(2x) = 2x + 1$ .

Now let's find $g(f(x))$ : $g(f(x)) = g(x+1) = 2(x+1) = 2x + 2$ .

We see that $2x + 1$ is not equal to $2x + 2$ , so $f(g(x))$ is not equal to $g(f(x))$ in general.

The statement says that if $f(x)$ is defined for all $x$ , then $f(f^{-1}(x)) = x$ .
This is the definition of an inverse function, but there's a catch!

The inverse function, $f^{-1}(x)$ , must also be defined for all $x$ for this to be true.
If $f(x)$ doesn't have an inverse defined for all $x$ , then the statement isn't necessarily true.

12. b. even
13. b) false
14. b) false
15. b) false