Chapter 5: Q. 84 (page 270)

Find the inverse of the function $f (x) = \sqrt{r^{2} - x^{2}} 0 \leq x \leq r$

Short Answer

Expert verified

The inverse of the function $f (x) = \sqrt{r^{2} - x^{2}} 0 \leq x \leq r$ is $f^{- 1} (x) = \sqrt{r^{2} - x^{2}} 0 \leq x \leq r$

Step by step solution

Step 1. Given data

The given function is

$f (x) = \sqrt{r^{2} - x^{2}} 0 \leq x \leq r$

Step 2. interchanging variables

Replace $f (x)$ with y and interchange x and y

$f (x) = \sqrt{r^{2} - x^{2}} y = \sqrt{r^{2} - x^{2}} x = \sqrt{r^{2} - y^{2}}$

Step 3. The inverse of the function

Solve the equation for y

$x = \sqrt{r^{2} - y^{2}} x^{2} = r^{2} - y^{2} y^{2} = r^{2} - x^{2} y = \sqrt{r^{2} - x^{2}}$

replace ywith $f^{- 1} (x)$

$y = \sqrt{r^{2} - x^{2}} f^{- 1} (x) = \sqrt{r^{2} - x^{2}}$

Step 4. Verification

Determine $f (f^{- 1} (x))$

$f (f^{- 1} (x)) = f (\sqrt{r^{2} - x^{2}}) f (f^{- 1} (x)) = \sqrt{r^{2} - {(\sqrt{r^{2} - x^{2}})}^{2}} f (f^{- 1} (x)) = \sqrt{r^{2} - r^{2} + x^{2}} f (f^{- 1} (x)) = \sqrt{x^{2}} f (f^{- 1} (x)) = x$

Determine $f^{- 1} (f (x))$

$f^{- 1} (f (x)) = f^{- 1} (\sqrt{r^{2} - x^{2}}) f^{- 1} (f (x)) = \sqrt{r^{2} - {(\sqrt{r^{2} - x^{2}})}^{2}} f^{- 1} (f (x)) = \sqrt{r^{2} - r^{2} + x^{2}} f^{- 1} (f (x)) = \sqrt{x^{2}} f^{- 1} (f (x)) = = x$

$f (f^{- 1} (x)) = x & f^{- 1} (f (x)) = x$ so inverse function is correct

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Find the inverse of the function $f (x) = \sqrt{r^{2} - x^{2}} 0 \leq x \leq r$

Short Answer

Step by step solution

Step 1. Given data

Step 2. interchanging variables

Step 3. The inverse of the function

Step 4. Verification

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Step by step solution

Step 1. Given data

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