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Chapter 8: Problem 63

Find the inverse function \(f^{-1}\) of each function \(f\). Find the range of f and the domain and range of \(f^{-1}\). \(f(x)=-\tan (x+1)-3 ;-1-\frac{\pi}{2}

Short Answer

Expert verified
Inverse function is \( f^{-1}(y) = \arctan (-(y+3)) - 1 \); Range of f: \((-\infty, -3) \); Domain of \( f^{-1} \): \((-\

Step by step solution

01

Express y in terms of x

Start with the equation of the function by letting y = f(x). So, we have: \( y = -\tan (x+1) - 3 \).
02

Solve for x in terms of y

To find the inverse function, solve the equation for x. First, add 3 to both sides:\( y + 3 = -\tan (x+1) \)Then, multiply both sides by -1:\( -y - 3 = \tan (x+1) \)
03

Use the inverse tangent function

Apply the arctangent to both sides to isolate x:\( \arctan (-(y+3)) = x + 1 \)
04

Solve for x

Subtract 1 from both sides to solve for x:\( x = \arctan (-(y+3)) - 1 \) This gives us the inverse function:\( f^{-1}(y) = \arctan (-(y+3)) - 1 \)
05

Determine the range of the original function f

The range of the original function can be found by evaluating the outputs as x varies within its given domain:\(-1-\frac{\pi}{2} < x < \frac{\pi}{2}-1\). Evaluating at the extremes of this interval gives \( f(-1-\frac{\pi}{2}) = -\tan(-\frac{\pi}{2})-3 \) and \( f(\frac{\pi}{2}-1) = -\tan(\frac{\pi}{2})-3 \). Since the range of \ \ \tan(x) \ \ is all real numbers, the range is: \( (-\infty, -3) \)
06

Determine the domain of the inverse function

The domain of the inverse function f^{-1} is the range of the original function f. Thus, the domain of f^{-1}(y) is:\( Range(f) = (-\infty, -3) \)
07

Determine the range of the inverse function

Since the domain of the inverse function corresponds to the original input of the function, the range of f^{-1}(y) is the original domain of f(x), giving: \( Range(f^{-1}) = (-1-\frac{\pi}{2}, \frac{\pi}{2}-1) \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Range of a Function
The range of a function is the set of all possible output values it can produce. When dealing with functions involving trigonometric terms like \(\tan(x)\), it's crucial to understand their behavior. Given a function like \(-\tan(x+1)-3\), the range is found by determining the outputs as the input, x, spans the given domain. In this example, the domain is \(-1-\frac{\pi}{2} < x < \frac{\pi}{2} - 1\). Evaluating the function at the endpoints, while considering the behavior of the \(\tan\) function, results in a range of \((-\infty, -3)\). Always remember, the range of the tangent function is all real numbers, but shifts and constants applied to tangent can affect this.
Domain of a Function
The domain of a function represents all the input values for which the function is defined. For the function \(-\tan(x+1)-3\), the given domain is \(-1- \frac{\pi}{2} < x < \frac{\pi}{2} - 1\). This domain ensures the function avoids undefined points where \(\tan(x)\) would approach \(\frac{\pi}{2}\) or its multiples. When the inverse function is determined as \(f^{-1}(y) = \arctan(-(y+3))-1\), the domain of this inverse function corresponds to the range of the original function. Therefore, the domain of the inverse function in this example is \((-\infty, -3)\).
Arctangent
The arctangent function, also known as \(\tan^{ -1}(x)\) or \(\text{arctan}(x)\), is the inverse of the tangent function. It returns the angle whose tangent is a given number. For example, \(\arctan(y)\) gives the angle \(\theta\) such that \(\tan(\theta) = y\). When dealing with our example's inverse function \(f^{-1}(y) = \arctan(-(y+3))-1\), applying arctangent helps isolate x. This is crucial because it helps us express y in terms of x. The range of the arctangent function is \((- \frac{\pi}{2}, \frac{\pi}{2})\), meaning it outputs values within this interval. This property often makes it a valuable tool in finding inverse functions for expressions involving tangent.

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Most popular questions from this chapter

Establish each identity. $$ \sin (\alpha+\beta)+\sin (\alpha-\beta)=2 \sin \alpha \cos \beta $$

If \(f(x)=\frac{1}{2} e^{x-1}+3,\) find the domain of \(f^{-1}(x)\)

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Area of a Dodecagon Part I A regular dodecagon is a polygon with 12 sides of equal length. See the figure. (a) The area \(A\) of a regular dodecagon is given by the formula \(A=12 r^{2} \tan \frac{\pi}{12},\) where \(r\) is the apothem, which is a line segment from the center of the polygon that is perpendicular to a side. Find the exact area of a regular dodecagon whose apothem is 10 inches. (b) The area \(A\) of a regular dodecagon is also given by the formula \(A=3 a^{2} \cot \frac{\pi}{12},\) where \(a\) is the length of a side of the polygon. Find the exact area of a regular dodecagon if the length of a side is 15 centimeters.

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