Solving Nonlinear Inequalities, Finding Function Domains, and Graphing

Published on March 1, 2026

Functions

This document covers solving various nonlinear inequalities (quadratic, rational, absolute value), determining the domains of complex algebraic and radical functions, and sketching graphs of piecewise and standard functions.

Question

solve these questions with proper steps

Answer

This appears to be a set of mathematics problems from a Pre-Calculus assignment. I will solve the problems visible in the provided images, addressing Question 1 (Inequalities), Question 2 (Domains), Question 3 (Graphing), Question 4 (Piecewise Graphs), and Question 5 (Function Definition).

Question 1: Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set.

i.

Step 1: Rewrite the inequality to have 0 on one side.

Step 2: Find the roots of the corresponding quadratic equation .
We can factor the quadratic:

The roots (critical points) are

and

Step 3: Test intervals.
The critical points divide the number line into three intervals:

, and

. Since the parabola

opens upwards (because the leading coefficient

), the expression will be positive outside the roots and negative between the roots. We are looking for where

Test : (True)
Test : (False)
Test : (True)

Step 4: Write the solution in interval notation.
Since the inequality is

, the endpoints are included.
Solution:

Graph: A number line with closed circles at

and

, shading extending to the left from

and to the right from

ii.

Step 1: Rewrite the inequality.

Multiply by

and reverse the inequality sign:

Step 2: Analyze the expression.
Since

for all real

, then

.
Therefore,

.
Since

, the inequality

is always true for all real numbers

Step 3: Write the solution in interval notation.
Solution:

Graph: A number line entirely shaded, with arrows on both ends.

iii.

Step 1: Rewrite the inequality to have 0 on one side.

Step 2: Combine terms using a common denominator.

Step 3: Analyze the resulting inequality.
For the fraction to be positive, the numerator and denominator must have the same sign.
The numerator is

, which is always negative.
Therefore, the denominator

must also be negative for the overall fraction to be positive:

Note: The critical points are where the numerator is zero (none) or the denominator is zero ().

Step 4: Write the solution in interval notation.
Solution:

Graph: A number line with an open circle at

, shading extending to the left.

iv.

Step 1: Rewrite the inequality to have 0 on one side.

Step 2: Combine terms using a common denominator.

Step 3: Find the critical points.
The expression changes sign when the numerator or denominator is zero:
Numerator:

;

.
Denominator:

.
Critical points:

Step 4: Test intervals.
Intervals:

. We seek where the expression is negative

Test : (False)
Test : (True)
Test : (False)
Test : (True)

Step 5: Write the solution in interval notation.
Since the inequality is strict (

), the endpoints are excluded.
Solution:

Graph: A number line with open circles at

, shading in the region between

and

, and the region to the right of

v.

Step 1: Solve the absolute value inequality.
The inequality

is equivalent to

Step 2: Isolate .
Subtract 5 from all parts of the inequality:

Step 3: Write the solution in interval notation.
Solution:

Graph: A number line with open circles at

and

, shading the region between them.

vi.

This is a compound inequality.

Step 1: Isolate in the middle.
Subtract 7 from all parts:

Step 2: Divide all parts by 3:

Step 3: Write the solution in interval notation.
Since the inequality includes

, the endpoints are included.
Solution:

Graph: A number line with closed circles at

and

, shading the segment between them.

vii.

Step 1: Isolate the absolute value expression.

Step 2: Analyze the result.
Since the absolute value

must be non-negative (i.e.,

), multiplying by 7 gives

.
Since

, the statement

is true for all real numbers

Step 3: Write the solution in interval notation.
Solution:

Graph: A number line entirely shaded.

viii.

Step 1: Rewrite the inequality to have 0 on one side.

Step 2: Combine terms using a common denominator.

Multiply by

and reverse the inequality sign:

Step 3: Find the critical points.
Numerator roots:

and

.
Denominator root:

.
Critical points:

Step 4: Test intervals.
Intervals:

. We seek where the expression is negative (

Test : (True)
Test : (Note: )
(False)
Test : (Note: )
(True)
Test : (False)

Step 5: Write the solution in interval notation.
Since the inequality is strict (

), the endpoints are excluded.
Solution:

Graph: A number line with open circles at

, and

, shading the region between

and

, and the region between

and

Question 2: Find the Domains of following functions.

The domain of a function is the set of all possible input values (

) for which the function is defined. Restrictions usually arise from:

Division by zero.
Taking the square root (or even root) of a negative number.

i.

Division by zero is the only restriction. Set the denominator to zero:

The function is undefined at

.
Domain:

ii.

There are two restrictions:

The numerator requires (for the square root).
The denominator cannot be zero: .

Find the roots of the denominator:

Roots are

and

We must satisfy

AND

.
Since

, the condition

is automatically satisfied.
We must exclude

from the set

Domain:

iii.

Set the denominator to zero:

.
The function is undefined at

and

.
Domain:

iv.

The expression under the square root must be non-negative:

From problem 2(ii), we know the roots are

and

.
Since the parabola

opens upwards, the expression is

outside the roots.

Domain:

Question 3: Sketch the graph of the function by first making a table of values.

i.

Domain: Requires

, so

. Domain is

			Point

Sketch: The graph starts at the origin

and extends to the left and up, resembling the square root function reflected across the y-axis.

ii.

We consider cases based on the definition of

:
Case 1: If

.
Case 2: If

		Point


(Boundary)

Sketch: For

, the graph is the line

(the x-axis). At

, the value is 0. For

, the graph is the line

. The graph is a single ray along the x-axis for

, meeting the line

at the origin.

iii.

This is the standard cubic function

shifted down 8 units.

			Point

Sketch: A smooth, increasing curve passing through

and

Question 4: Sketch the graph of the piecewise defined function.

i.

Piece 1: for (Open circle at )
At

. Boundary point is

(open).
Test

. Point

Piece 2: for (Closed circle at )
At

. Boundary point is

(closed).
Test

. Point

.
Test

. Point

Sketch: A ray with slope 2 ending at an open circle at

. A ray starting from a closed circle at

with slope

. There is a jump discontinuity at

ii.

Note: Assuming simplifies to for .

Piece 1: for (Closed circle at )
At

. Boundary point is

(closed).
Test

. Point

.
Test

. Point

Piece 2: for (Open circle at )
At

: Limit approaches 1. Boundary point is

(open).
For

, the graph is the horizontal line

Sketch: A segment of a downward-opening parabola (

) that includes the point

. At

, the function jumps up to an open circle at

and continues horizontally as

for all

Question 5: Determine whether the equation defines as a function of .

For

to be a function of

, every input

must correspond to exactly one output

. This often means solving for

and checking if we get one or two results.

i.

Solve for

For any given

yields a single non-negative value, so

yields a single non-positive value.
Result: Yes, this defines

as a function of

ii.

This is the equation of a circle centered at

with radius

.
Solve for

For any

such that

(i.e.,

), there will be two corresponding

values (one using

and one using

).
Result: No, this does not define

as a function of

(it fails the vertical line test).

iii.

Solve for

This equation requires

. For any positive

, there are two real values of

(one positive, one negative). For example, if

.
Result: No, this does not define

as a function of

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Solving Nonlinear Inequalities, Finding Function Domains, and Graphing

Question

Answer

Question 1: Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set.

i.

ii.

iii.

iv.

v.

vi.

vii.

viii.

Question 2: Find the Domains of following functions.

i.

ii.

iii.

iv.

Question 3: Sketch the graph of the function by first making a table of values.

i.

ii.

iii.

Question 4: Sketch the graph of the piecewise defined function.

i.

ii.

Question 5: Determine whether the equation defines as a function of .

i.

ii.

iii.

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