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Given $f(x) = 1 + oot{x}$, what are the domain and range of $f^{-1}(x)$? A - HSC - SSCE Mathematics Extension 1 - Question 2 - 2020 - Paper 1

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Given-$f(x)-=-1-+--oot{x}$,-what-are-the-domain-and-range-of-$f^{-1}(x)$?--A-HSC-SSCE Mathematics Extension 1-Question 2-2020-Paper 1.png

Given $f(x) = 1 + oot{x}$, what are the domain and range of $f^{-1}(x)$? A. $x \geq 0, \; y \geq 0$ B. $x \geq 0, \; y \geq 1$ C. $x \geq 1, \; y \geq 0$ D. $x \ge... show full transcript

Worked Solution & Example Answer:Given $f(x) = 1 + oot{x}$, what are the domain and range of $f^{-1}(x)$? A - HSC - SSCE Mathematics Extension 1 - Question 2 - 2020 - Paper 1

Step 1

Determine the function and its inverse

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Answer

The function is defined as f(x)=1+xf(x) = 1 + \sqrt{x}. To find the domain and range of the inverse, we first need to derive the inverse function.

Starting from: y=1+xy = 1 + \sqrt{x} To find the inverse, solve for xx: y−1=xy - 1 = \sqrt{x} Squaring both sides gives: x=(y−1)2x = (y - 1)^2 Thus, the inverse function is: f−1(y)=(y−1)2f^{-1}(y) = (y - 1)^2

Step 2

Find the domain of the inverse function

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The expression (y−1)2(y - 1)^2 is defined for all real numbers yy. However, since x\sqrt{x} requires x≥0x \geq 0, we must consider the range of f(x)f(x) which is y≥1y \geq 1 after the adjustments with 11. Thus the domain of f−1(y)f^{-1}(y) is: y≥1y \geq 1

Step 3

Find the range of the inverse function

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Answer

For the inverse function f−1(y)=(y−1)2f^{-1}(y) = (y - 1)^2, since (y−1)2≥0(y - 1)^2 \geq 0 when y≥1y \geq 1, the range is: x≥0x \geq 0

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