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 $\left.f f^{-1}(x)\right)=x$ . a) True b) false

Studdy Solution

STEP 1

What is this asking? We're plugging $\pi$ into $\cos(x)$ , and then plugging that result into $g(x) = \sqrt{10 + x}$ . Watch out! Make sure to evaluate the functions from the inside out!
Also, remember what $\cos(\pi)$ equals.

STEP 2

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

STEP 3

Alright, so we start with our function $f(x) = \cos(x)$ .
We want to find $f(\pi)$ , which just means plugging in $\pi$ for $x$ .
So, we get $f(\pi) = \cos(\pi)$ .

STEP 4

Now, remember the unit circle!
At $\pi$ radians, the x-coordinate is -1, which is precisely the value of $\cos(\pi)$ .
So, $f(\pi) = \cos(\pi) = \mathbf{-1}$ .
Awesome!

STEP 5

We know that $g(x) = \sqrt{10 + x}$ and we just found that $f(\pi) = \mathbf{-1}$ .
This means $g(f(\pi))$ is the same as $g(\mathbf{-1})$ .

STEP 6

Let's plug in $\mathbf{-1}$ for $x$ in $g(x)$ .
We get $g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})}$ .

STEP 7

Inside the square root, we have $10 + (\mathbf{-1}) = 10 - 1 = \mathbf{9}$ .
So, our expression becomes $\sqrt{\mathbf{9}}$ .

STEP 8

Finally, the square root of $\mathbf{9}$ is $\mathbf{3}$ !
So, $g(f(\pi)) = g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})} = \sqrt{\mathbf{9}} = \mathbf{3}$ .
We did it!

STEP 9

The value of $g(f(\pi))$ is $\mathbf{3}$ .

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Studdy Solution

STEP 1

What is this asking? We're plugging $\pi$ into $\cos(x)$ , and then plugging that result into $g(x) = \sqrt{10 + x}$ . Watch out! Make sure to evaluate the functions from the inside out!
Also, remember what $\cos(\pi)$ equals.

STEP 2

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

STEP 3

Alright, so we start with our function $f(x) = \cos(x)$ .
We want to find $f(\pi)$ , which just means plugging in $\pi$ for $x$ .
So, we get $f(\pi) = \cos(\pi)$ .

STEP 4

Now, remember the unit circle!
At $\pi$ radians, the x-coordinate is -1, which is precisely the value of $\cos(\pi)$ .
So, $f(\pi) = \cos(\pi) = \mathbf{-1}$ .
Awesome!

STEP 5

We know that $g(x) = \sqrt{10 + x}$ and we just found that $f(\pi) = \mathbf{-1}$ .
This means $g(f(\pi))$ is the same as $g(\mathbf{-1})$ .

STEP 6

Let's plug in $\mathbf{-1}$ for $x$ in $g(x)$ .
We get $g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})}$ .

STEP 7

Inside the square root, we have $10 + (\mathbf{-1}) = 10 - 1 = \mathbf{9}$ .
So, our expression becomes $\sqrt{\mathbf{9}}$ .

STEP 8

Finally, the square root of $\mathbf{9}$ is $\mathbf{3}$ !
So, $g(f(\pi)) = g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})} = \sqrt{\mathbf{9}} = \mathbf{3}$ .
We did it!

STEP 9

The value of $g(f(\pi))$ is $\mathbf{3}$ .

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Studdy Solution

STEP 1

What is this asking? We're plugging π\piπ into cos⁡(x)\cos(x)cos(x), and then plugging *that* result into g(x)=10+xg(x) = \sqrt{10 + x}g(x)=10+x​. Watch out! Make sure to evaluate the functions from the inside out!Also, remember what cos⁡(π)\cos(\pi)cos(π) equals.

STEP 2

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

STEP 3

Alright, so we **start** with our function f(x)=cos⁡(x)f(x) = \cos(x)f(x)=cos(x).We want to find f(π)f(\pi)f(π), which just means plugging in π\piπ for xxx.So, we get f(π)=cos⁡(π)f(\pi) = \cos(\pi)f(π)=cos(π).

STEP 4

Now, remember the **unit circle**!At π\piπ radians, the x-coordinate is **-1**, which is precisely the value of cos⁡(π)\cos(\pi)cos(π).So, f(π)=cos⁡(π)=−1f(\pi) = \cos(\pi) = \mathbf{-1}f(π)=cos(π)=−1.Awesome!

STEP 5

We know that g(x)=10+xg(x) = \sqrt{10 + x}g(x)=10+x​ and we just found that f(π)=−1f(\pi) = \mathbf{-1}f(π)=−1.This means g(f(π))g(f(\pi))g(f(π)) is the same as g(−1)g(\mathbf{-1})g(−1).

STEP 6

Let's **plug** in −1\mathbf{-1}−1 for xxx in g(x)g(x)g(x).We get g(−1)=10+(−1)g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})}g(−1)=10+(−1)​.

STEP 7

Inside the square root, we have 10+(−1)=10−1=910 + (\mathbf{-1}) = 10 - 1 = \mathbf{9}10+(−1)=10−1=9.So, our expression becomes 9\sqrt{\mathbf{9}}9​.

STEP 8

Finally, the square root of 9\mathbf{9}9 is 3\mathbf{3}3!So, g(f(π))=g(−1)=10+(−1)=9=3g(f(\pi)) = g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})} = \sqrt{\mathbf{9}} = \mathbf{3}g(f(π))=g(−1)=10+(−1)​=9​=3.We did it!

STEP 9

The value of g(f(π))g(f(\pi))g(f(π)) is 3\mathbf{3}3.

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What is this asking? We're plugging $\pi$ into $\cos(x)$ , and then plugging that result into $g(x) = \sqrt{10 + x}$ . Watch out! Make sure to evaluate the functions from the inside out!
Also, remember what $\cos(\pi)$ equals.

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

Alright, so we start with our function $f(x) = \cos(x)$ .
We want to find $f(\pi)$ , which just means plugging in $\pi$ for $x$ .
So, we get $f(\pi) = \cos(\pi)$ .

Now, remember the unit circle!
At $\pi$ radians, the x-coordinate is -1, which is precisely the value of $\cos(\pi)$ .
So, $f(\pi) = \cos(\pi) = \mathbf{-1}$ .
Awesome!

We know that $g(x) = \sqrt{10 + x}$ and we just found that $f(\pi) = \mathbf{-1}$ .
This means $g(f(\pi))$ is the same as $g(\mathbf{-1})$ .

Let's plug in $\mathbf{-1}$ for $x$ in $g(x)$ .
We get $g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})}$ .

Inside the square root, we have $10 + (\mathbf{-1}) = 10 - 1 = \mathbf{9}$ .
So, our expression becomes $\sqrt{\mathbf{9}}$ .

Finally, the square root of $\mathbf{9}$ is $\mathbf{3}$ !
So, $g(f(\pi)) = g(\mathbf{-1}) = \sqrt{10 + (\mathbf{-1})} = \sqrt{\mathbf{9}} = \mathbf{3}$ .
We did it!

The value of $g(f(\pi))$ is $\mathbf{3}$ .