Example 1: Find the Range of function f
defined by
Solution to Example 1

Let us first write the given function as an equation as follows

Solve the above equation for x
y (2x  3) = x + 1
2 y x  3 y = x + 1
2 y x  x = 3 y + 1
x (2 y  1) = 3 y + 1
x = (3 y + 1) / (2 y  1)

The above expression of x in terms of y shows that x is real for all real values of y except 1/2 since y = 1 / 2 will make the denominator 2 y  1 = 0. Hence the range of f, which is the set of all possible values of y, is given by
(∞ , 1 / 2) ∪ (1 / 2 , +∞)
See graph below of function f given above and compare range found and that of the graph.
Matched Problem 1: Find the range of
function f defined by
Answer at the bottom of the page.
Example 2: Find the Range of function f
defined by
Solution to Example 2

Write the given function as an equation

Rewrite the above equation for x in standard form and solve using quadratic formulae
y x^{2}  x  9 y  2 = 0

Find the discriminant to the above equation
Δ = (1)^{2}  4 y (9 y  2) = 36 y^{2} + 8 y + 1

Using the quadratic formulas, the above equation gives the solutions

The solutions x_{1,2} are real if 36 y^{2} + 8 y + 1 ≥ 0 and y ≠0. Hence we need to solve the inequality
36 y^{2} + 8 y + 1 ≥ 0

The discriminant of 36 y^{2} + 8 y + 1 is equal to
8^{2}  4(36)(1) =  80

Since the discriminant is negative, one test value shows that 36 y^{2} + 8 y + 1 is always positive. For y = 0, we need to set y= 0 in the equation y = (x + 2) / (x^{2}  9) and solve it
(x + 2) / (x^{2}  9) = 0

gives solution x = 2 and therefore y = 0 is also in the range of f. Hence the range of f is given by
( ∞ , +∞ )
See graph below of function f given above and compare range found and that of the graph.
Matched Problem 2: Find the range of
function f defined by
Answer at the bottom of the page.
Example 3: Find the Range of function f
defined by
Solution to Example 3

Write the above function as an equation.

Rewrite the above as follows.
y x^{2}  x + y  2 = 0

Solve for x using the quadratic formulae.

The above solutions are real if the radicand is not negative and y not equal to 0. Hence we need to solve the inequality
1  4 y^{2} + 8y ≥ 0

The solution set to the above inequality is
1  √5 / 2 ≤ y ≤ 1 + √5 / 2 with y = 0 excluded.

But if we set y to 0 in the first equation, we obtain
0 = (x + 2) / (x^{2} + 1)

which gives x =  2 and hence y = 0 is also included in the range of f for now. Hence the range of f is given by the interval
[ 1  √5 / 2 , 1 + √5 / 2 ]
See graph below of function f given above and compare range found and that of the graph.
Matched Problem 3: Find the range of
function f defined by
Answer at the bottom of the page.
Example 4: Find the Range of function f
defined by
Solution to Example 4

Write the function as an equation

rewrite as a quadratic equation in x
x^{2}( y  1 ) = 2  y

Solve for x

The above solutions are real if the radicand is not negative. Hence we need to solve the inequality
(2  y) / (y  1) ≥ 0

whose solution set is given by
1 < y ≤ 2

The range of the given function is given by the interval
(1 , 2]
See graph below of function f given above and compare range found and that of the graph.
Matched Problem 4: Find the range of
function f defined by
Answer at the bottom of the page.
Example 5: Find the Range of function f defined by
Solution to Example 5

Write the function as an equation

rewrite as a quadratic equation in x
x^{2}y = y + 1

Solve for x

for the solutions to be real, the radicand must be non negative. Hence
(y + 1) / y ≥ 0

The solution set to the above inequality is
(∞ 1 ] ∪ (0 , +∞)

which is also the range of the given function.
See graph below of function f given above and compare range found and that of the graph.
Matched Problem 5: Find the range of
function f defined by
Answer at the bottom of the page.
Example 6: Find the Range of function f
defined by
Solution to Example 6

Write the function as an equation

rewrite the above as a quadratic function
2 x ^{2}  y x  y  1 = 0

solve for x

the above solutions are real if the radicand is not negative. Hence we need to solve the inequality
y^{2} + 8 y + 8 ≥ 0

whose solution set is given by the interval
(∞ ,  4  2√2 ] ∪ [ 4 + 2√2 , +∞)

which is also the range of the given function
See graph below of function f given above and compare range found and that of the graph.
Matched Problem 6: Find the range of
function f defined by
Answer at the bottom of the page.
Answers to matched problems
1. (∞ , 1) ∪ (1 , +∞)
2. (∞ , +∞)
3. [ (2  √10) /12 , (2 + √10) /12 ]
4. (1/2 , 5]
5. (∞ 0 ) ∪ [ 3/4 , +∞)
6. (∞ , 42√6 ] ∪ [4+2√6 , +∞)
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