Math / Algebra

Question11. Given the functions $f(x)=\cos x$ and $g(x)=\sqrt{10+x}$ , determine the value of $g(f(\pi))$ . a.
3 b. 9 c. $\sqrt{11}$ d. $\sqrt{\pi}$
12. 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

Studdy Solution

STEP 1

What is this asking? We need to plug $\pi$ into $f(x)$ , then plug that result into $g(x)$ . Watch out! Make sure to follow the order of operations and work from the inside out!

STEP 2

1. Evaluate $f(\pi)$
2. Evaluate $g(f(\pi))$

STEP 3

Alright, let's kick things off by finding $f(\pi)$ .
We're given $f(x) = \cos x$ , so we need to substitute $\pi$ for $x$ .

STEP 4

That gives us $f(\pi) = \cos \pi$ .
Remember, the cosine of $\pi$ is \-1!
So, $f(\pi) = \textbf{-1}$ .
This is a key result we'll use in the next step.

STEP 5

Now, we need to find $g(f(\pi))$ .
We just figured out that $f(\pi) = \textbf{-1}$ , so we're really looking for $g(\textbf{-1})$ .

STEP 6

We know that $g(x) = \sqrt{10 + x}$ .
Let's plug in our value for $f(\pi)$ , which is $\textbf{-1}$ , into $g(x)$ .

STEP 7

So, we have $g(\textbf{-1}) = \sqrt{10 + (\textbf{-1})}$ .
This simplifies to $g(\textbf{-1}) = \sqrt{10 - 1} = \sqrt{9}$ .

STEP 8

And finally, $\sqrt{9} = \textbf{3}$ !

STEP 9

So, $g(f(\pi)) = \textbf{3}$ , which means the answer is a.

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Studdy Solution

STEP 1

What is this asking? We need to plug $\pi$ into $f(x)$ , then plug that result into $g(x)$ . Watch out! Make sure to follow the order of operations and work from the inside out!

STEP 2

1. Evaluate $f(\pi)$
2. Evaluate $g(f(\pi))$

STEP 3

Alright, let's kick things off by finding $f(\pi)$ .
We're given $f(x) = \cos x$ , so we need to substitute $\pi$ for $x$ .

STEP 4

That gives us $f(\pi) = \cos \pi$ .
Remember, the cosine of $\pi$ is \-1!
So, $f(\pi) = \textbf{-1}$ .
This is a key result we'll use in the next step.

STEP 5

Now, we need to find $g(f(\pi))$ .
We just figured out that $f(\pi) = \textbf{-1}$ , so we're really looking for $g(\textbf{-1})$ .

STEP 6

We know that $g(x) = \sqrt{10 + x}$ .
Let's plug in our value for $f(\pi)$ , which is $\textbf{-1}$ , into $g(x)$ .

STEP 7

So, we have $g(\textbf{-1}) = \sqrt{10 + (\textbf{-1})}$ .
This simplifies to $g(\textbf{-1}) = \sqrt{10 - 1} = \sqrt{9}$ .

STEP 8

And finally, $\sqrt{9} = \textbf{3}$ !

STEP 9

So, $g(f(\pi)) = \textbf{3}$ , which means the answer is a.

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Studdy Solution

STEP 1

What is this asking? We need to plug π\piπ into f(x)f(x)f(x), then plug *that* result into g(x)g(x)g(x). Watch out! Make sure to follow the order of operations and work from the inside out!

STEP 2

1. Evaluate f(π)f(\pi)f(π)2. Evaluate g(f(π))g(f(\pi))g(f(π))

STEP 3

Alright, let's **kick things off** by finding f(π)f(\pi)f(π).We're given f(x)=cos⁡xf(x) = \cos xf(x)=cosx, so we need to **substitute** π\piπ for xxx.

STEP 4

That gives us f(π)=cos⁡πf(\pi) = \cos \pif(π)=cosπ.Remember, the **cosine of π\piπ** is **\-1**!So, f(π)=-1f(\pi) = \textbf{-1}f(π)=-1.This is a **key result** we'll use in the next step.

STEP 5

Now, we need to find g(f(π))g(f(\pi))g(f(π)).We just figured out that f(π)=-1f(\pi) = \textbf{-1}f(π)=-1, so we're really looking for g(-1)g(\textbf{-1})g(-1).

STEP 6

We know that g(x)=10+xg(x) = \sqrt{10 + x}g(x)=10+x​.Let's **plug in** our value for f(π)f(\pi)f(π), which is -1\textbf{-1}-1, into g(x)g(x)g(x).

STEP 7

So, we have g(-1)=10+(-1)g(\textbf{-1}) = \sqrt{10 + (\textbf{-1})}g(-1)=10+(-1)​.This simplifies to g(-1)=10−1=9g(\textbf{-1}) = \sqrt{10 - 1} = \sqrt{9}g(-1)=10−1​=9​.

STEP 8

And finally, 9=3\sqrt{9} = \textbf{3}9​=3!

STEP 9

So, g(f(π))=3g(f(\pi)) = \textbf{3}g(f(π))=3, which means the answer is **a**.

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What is this asking? We need to plug $\pi$ into $f(x)$ , then plug that result into $g(x)$ . Watch out! Make sure to follow the order of operations and work from the inside out!

1. Evaluate $f(\pi)$
2. Evaluate $g(f(\pi))$

Alright, let's kick things off by finding $f(\pi)$ .
We're given $f(x) = \cos x$ , so we need to substitute $\pi$ for $x$ .

That gives us $f(\pi) = \cos \pi$ .
Remember, the cosine of $\pi$ is \-1!
So, $f(\pi) = \textbf{-1}$ .
This is a key result we'll use in the next step.

Now, we need to find $g(f(\pi))$ .
We just figured out that $f(\pi) = \textbf{-1}$ , so we're really looking for $g(\textbf{-1})$ .

We know that $g(x) = \sqrt{10 + x}$ .
Let's plug in our value for $f(\pi)$ , which is $\textbf{-1}$ , into $g(x)$ .

So, we have $g(\textbf{-1}) = \sqrt{10 + (\textbf{-1})}$ .
This simplifies to $g(\textbf{-1}) = \sqrt{10 - 1} = \sqrt{9}$ .

And finally, $\sqrt{9} = \textbf{3}$ !

So, $g(f(\pi)) = \textbf{3}$ , which means the answer is a.