Math / Trigonometry

QuestionWhat is the range of $y=\sec ^{-1}(x) ?$ $\left[-\frac{\pi}{2}, 0\right)$ and $\left(0, \frac{\pi}{2}\right]$ $\left[-\frac{\pi}{2}, 0\right]$ and $\left[0, \frac{\pi}{2}\right]$ $\left[0, \frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2}, \pi\right]$ $\left[0, \frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2}, \pi\right]$

Studdy Solution

STEP 1

What is this asking? What are all the possible output values when we plug any allowed $x$ into the inverse secant function, $\sec^{-1}(x)$ ? Watch out! Remember the inverse secant function has a restricted domain and range, unlike the regular secant function!
Don't mix them up!

STEP 2

1. Define the range of inverse secant.
2. Relate secant and cosine.
3. Analyze the range.

STEP 3

Alright, let's define our function!
The inverse secant, written as $y = \sec^{-1}(x)$ , is the inverse function of the secant function.
This means it essentially "undoes" what the secant function does.
So, if $\sec(\theta) = x$ , then $\sec^{-1}(x) = \theta$ .

STEP 4

But there's a catch!
The secant function isn't one-to-one, meaning different angles can give you the same secant value.
To make the inverse secant a function, we have to restrict its range.
This is a standard restriction, not something we make up on the fly!

STEP 5

The range of $\sec^{-1}(x)$ is $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$ .
Notice that $\frac{\pi}{2}$ is carefully excluded!
This is because $\sec(\frac{\pi}{2})$ is undefined.

STEP 6

Let's think about how secant and cosine are related.
Remember, the secant of an angle is the reciprocal of the cosine of that same angle.
Mathematically, this means $\sec(\theta) = \frac{1}{\cos(\theta)}$ .

STEP 7

Since $\sec(\theta) = x$ , we can also write $\frac{1}{\cos(\theta)} = x$ .
This means $\cos(\theta) = \frac{1}{x}$ .
This relationship is super helpful because it connects the inverse secant to the more familiar cosine function.

STEP 8

We know the range of $\sec^{-1}(x)$ is $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$ .
Let's think about what this means for the cosine.
When $\theta$ is between $0$ and $\frac{\pi}{2}$ (excluding $\frac{\pi}{2}$ ), $\cos(\theta)$ is positive.
When $\theta$ is between $\frac{\pi}{2}$ and $\pi$ (excluding $\frac{\pi}{2}$ ), $\cos(\theta)$ is negative.

STEP 9

This makes sense!
Since $\cos(\theta) = \frac{1}{x}$ , $x$ can be any number greater than or equal to 1, or any number less than or equal to -1.
This corresponds perfectly to the range of $\sec^{-1}(x)$ that we defined earlier.

STEP 10

The range of $y = \sec^{-1}(x)$ is $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$ .
So the correct answer is $\left[0, \frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2}, \pi\right]$ .

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Studdy Solution

STEP 1

What is this asking? What are all the possible output values when we plug any allowed $x$ into the inverse secant function, $\sec^{-1}(x)$ ? Watch out! Remember the inverse secant function has a restricted domain and range, unlike the regular secant function!
Don't mix them up!

STEP 2

1. Define the range of inverse secant.
2. Relate secant and cosine.
3. Analyze the range.

STEP 3

Alright, let's define our function!
The inverse secant, written as $y = \sec^{-1}(x)$ , is the inverse function of the secant function.
This means it essentially "undoes" what the secant function does.
So, if $\sec(\theta) = x$ , then $\sec^{-1}(x) = \theta$ .

STEP 4

But there's a catch!
The secant function isn't one-to-one, meaning different angles can give you the same secant value.
To make the inverse secant a function, we have to restrict its range.
This is a standard restriction, not something we make up on the fly!

STEP 5

The range of $\sec^{-1}(x)$ is $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$ .
Notice that $\frac{\pi}{2}$ is carefully excluded!
This is because $\sec(\frac{\pi}{2})$ is undefined.

STEP 6

Let's think about how secant and cosine are related.
Remember, the secant of an angle is the reciprocal of the cosine of that same angle.
Mathematically, this means $\sec(\theta) = \frac{1}{\cos(\theta)}$ .

STEP 7

Since $\sec(\theta) = x$ , we can also write $\frac{1}{\cos(\theta)} = x$ .
This means $\cos(\theta) = \frac{1}{x}$ .
This relationship is super helpful because it connects the inverse secant to the more familiar cosine function.

STEP 8

STEP 9

This makes sense!
Since $\cos(\theta) = \frac{1}{x}$ , $x$ can be any number greater than or equal to 1, or any number less than or equal to -1.
This corresponds perfectly to the range of $\sec^{-1}(x)$ that we defined earlier.

STEP 10

The range of $y = \sec^{-1}(x)$ is $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$ .
So the correct answer is $\left[0, \frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2}, \pi\right]$ .

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Studdy Solution

STEP 1

STEP 2

1. Define the range of inverse secant.2. Relate secant and cosine.3. Analyze the range.

STEP 3

STEP 4

But there's a catch!The secant function isn't one-to-one, meaning different angles can give you the same secant value.To make the inverse secant a *function*, we have to **restrict its range**.This is a standard restriction, not something we make up on the fly!

STEP 5

The **range** of sec⁡−1(x)\sec^{-1}(x)sec−1(x) is [0,π2)[0, \frac{\pi}{2})[0,2π​) and (π2,π](\frac{\pi}{2}, \pi](2π​,π].Notice that π2\frac{\pi}{2}2π​ is carefully excluded!This is because sec⁡(π2)\sec(\frac{\pi}{2})sec(2π​) is undefined.

STEP 6

Let's think about how secant and cosine are related.Remember, the secant of an angle is the reciprocal of the cosine of that same angle.Mathematically, this means sec⁡(θ)=1cos⁡(θ)\sec(\theta) = \frac{1}{\cos(\theta)}sec(θ)=cos(θ)1​.

STEP 7

Since sec⁡(θ)=x\sec(\theta) = xsec(θ)=x, we can also write 1cos⁡(θ)=x\frac{1}{\cos(\theta)} = xcos(θ)1​=x.This means cos⁡(θ)=1x\cos(\theta) = \frac{1}{x}cos(θ)=x1​.This relationship is super helpful because it connects the inverse secant to the more familiar cosine function.

STEP 8

STEP 9

This makes sense!Since cos⁡(θ)=1x\cos(\theta) = \frac{1}{x}cos(θ)=x1​, xxx can be *any* number greater than or equal to 1, or any number less than or equal to -1.This corresponds perfectly to the range of sec⁡−1(x)\sec^{-1}(x)sec−1(x) that we defined earlier.

STEP 10

The range of y=sec⁡−1(x)y = \sec^{-1}(x)y=sec−1(x) is [0,π2)[0, \frac{\pi}{2})[0,2π​) and (π2,π](\frac{\pi}{2}, \pi](2π​,π].So the correct answer is [0,π2)\left[0, \frac{\pi}{2}\right)[0,2π​) and (π2,π]\left(\frac{\pi}{2}, \pi\right](2π​,π].

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1. Define the range of inverse secant.
2. Relate secant and cosine.
3. Analyze the range.

But there's a catch!
The secant function isn't one-to-one, meaning different angles can give you the same secant value.
To make the inverse secant a function, we have to restrict its range.
This is a standard restriction, not something we make up on the fly!

The range of $\sec^{-1}(x)$ is $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$ .
Notice that $\frac{\pi}{2}$ is carefully excluded!
This is because $\sec(\frac{\pi}{2})$ is undefined.

Let's think about how secant and cosine are related.
Remember, the secant of an angle is the reciprocal of the cosine of that same angle.
Mathematically, this means $\sec(\theta) = \frac{1}{\cos(\theta)}$ .

Since $\sec(\theta) = x$ , we can also write $\frac{1}{\cos(\theta)} = x$ .
This means $\cos(\theta) = \frac{1}{x}$ .
This relationship is super helpful because it connects the inverse secant to the more familiar cosine function.

This makes sense!
Since $\cos(\theta) = \frac{1}{x}$ , $x$ can be any number greater than or equal to 1, or any number less than or equal to -1.
This corresponds perfectly to the range of $\sec^{-1}(x)$ that we defined earlier.

The range of $y = \sec^{-1}(x)$ is $[0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi]$ .
So the correct answer is $\left[0, \frac{\pi}{2}\right)$ and $\left(\frac{\pi}{2}, \pi\right]$ .