Answer:
Given a function f(x) = y, to find the range of f(x) is to solve for the value of x using the same function
If y is in denominator, the range of the function is all values of y wherein the denominator is ≠ 0
If y is in numerator or the function is an equation, then the range is all Real numbers ℝ
If the function is a radical and y is in the radicand, then the range is all values of y where the radicand > - 1
\:
1. y = 3x + 2
\implies x = \frac{y-2}{3}⟹x=3y−2
Since y is in numerator, then the range is all Real numbers ℝ
Range = ℝ
\:
2. x + y = 8
\implies x = 8 - y⟹x=8−y
Since the function is a linear function, then the range is all Real numbers ℝ
Range = ℝ
\:
3. y = 5x - 1
\implies x = \frac{y+1}{5}⟹x=5y+1
Since y is in numerator, then the range is all Real numbers ℝ
Range = ℝ
\:
4. y = 3x²
\implies \sqrt{3y}⟹3y
Since the function is a radical, then the range is all values of y where √(3y) > - 1
\:
From the inequality √(3y) > - 1, we can already solve for the range
\implies \sqrt{3y}\gt - 1⟹3y>−1
\implies {\sqrt{3y}}^{2}\gt - 1^2⟹3y2>−12
\implies 3y\gt 1⟹3y>1
\therefore y\gt\frac{1}{3}∴y>31
Range = y > ⅓
\:
5. y = 2x - ½
\implies x = \frac{2y+1}{4}⟹x=42y+1
Since y is in numerator, then the range is all Real numbers ℝ
Range = ℝ
\:
6. x - 2y = 6
\implies x = 6 + 2y⟹x=6+2y
Since the function is a linear function, then the range is all Real numbers ℝ
Range = ℝ
\:
7. y = (x² - 1) / 1
\implies x = \sqrt{y+1}⟹x=y+1
Since the function is a radical, then the range is all values of y where √(y+1) > - 1
\:
From the inequality √(y+1) > - 1, we can already solve for the range
\implies \sqrt{y+1}\gt - 1⟹y+1>−1
\implies {\sqrt{y+1}}^{2}\gt - 1^2⟹y+12>−12
\implies y+1\gt 1⟹y+1>1
\therefore y\gt 0∴y>0
Range = y > 0
\:
8. x = y - 3
Since the function is a linear function, then the range is all Real numbers ℝ
Range = ℝ
\:
9. y = x² - 4x - 3
Since the function is a quadratic equation, then the range is all Real numbers ℝ
Range = ℝ
\:
10. y = (x - 1)(x + 1)
\implies x = \sqrt{y+1}⟹x=y+1
Since the function is a radical, then the range is all values of y where √(y+1) > - 1
\:
From the inequality √(y+1) > - 1, we can already solve for the range
\implies \sqrt{y+1}\gt - 1⟹y+1>−1
\implies {\sqrt{y+1}}^{2}\gt - 1^2⟹y+12>−12
\implies y+1\gt 1⟹y+1>1
\therefore y\gt 0∴y>0
Range = y > 0
