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Chapter 0: Problem 20

Find \(f^{-1}(a)\) for the function \(f\) and the real number \(a\). $$ f(x)=2+\tan \left(\frac{\pi x}{2}\right), \quad-1

Short Answer

Expert verified
The short answer is: \(f^{-1}(2) = 0\).

Step by step solution

01

Swap x and y in the equation

We start by swapping x and y in the equation, so we get: $$ x = 2 + \tan\left(\frac{\pi y}{2}\right) $$
02

Isolate y in the equation

Now, we will isolate y by applying these steps: Step 2.1: Subtract 2 from both sides: $$ x - 2 = \tan\left(\frac{\pi y}{2}\right) $$ Step 2.2: Apply the arctangent function to both sides: $$ \tan^{-1}(x - 2) = \tan^{-1}\left(\tan(\frac{\pi y}{2})\right) $$ Step 2.3: Use the property that the arctangent and tangent functions are inverse functions: $$ \tan^{-1}(x - 2) = \frac{\pi y}{2} $$ Step 2.4: Multiply by 2 and divide by pi to isolate y: $$ y = \frac{2}{\pi}\tan^{-1}(x - 2) $$ The inverse function is given by: $$ f^{-1}(x) = \frac{2}{\pi}\tan^{-1}(x - 2) $$
03

Evaluate the inverse function at the given value of a

Since we are given that \(a = 2\), we will plug in this value into the inverse function and solve for x: $$ f^{-1}(2) = \frac{2}{\pi}\tan^{-1}(2 - 2) = \frac{2}{\pi}\tan^{-1}(0) $$ Now, we know that \(\tan^{-1}(0) = 0\), so: $$ f^{-1}(2) = \frac{2}{\pi}(0) = 0 $$ Thus, the value of \(f^{-1}(a)\), when \(a = 2\), is \(f^{-1}(2) = 0\).

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