Determine the domain of the real-valued function $f(x) = \sqrt{x^2 - 2x - 15} + \ln(10 - 2x)$ by setting up and solving the coupled system of inequalities required for both the radicand and logarithmic argument. Express the resulting domain in interval notation and describe the exact graph of this solution set on a real number line.

Final answer

Domain in interval notation: $(-\infty, -3]$ ; on the number line, a solid closed circle at $x = -3$ with a shaded ray extending indefinitely to the left.

Steps

Condition for the square root expression

To find the domain of the real-valued function $f(x) = \sqrt{x^2 - 2x - 15} + \ln(10 - 2x)$ , both components of the function must be defined simultaneously over the real numbers. This requires setting up and solving a system of two inequalities.

For the square root $\sqrt{x^2 - 2x - 15}$ to produce real values, the radicand must be non-negative: $x^2 - 2x - 15 \ge 0$

Factor the quadratic expression: $(x - 5)(x + 3) \ge 0$

The critical values are $x = -3$ and $x = 5$ . Testing the signs in the intervals determined by these roots:

For $x < -3$ (e.g., $x = -4$ ): $(-9)(-1) > 0$ (positive)
For $-3 < x < 5$ (e.g., $x = 0$ ): $(-5)(3) < 0$ (negative)
For $x > 5$ (e.g., $x = 6$ ): $(1)(9) > 0$ (positive)

Including the endpoints where the product is zero gives: $x \in (-\infty, -3] \cup [5, \infty)$ ](streamdown:incomplete-link)

Condition for the natural logarithm expression

For the natural logarithm $\ln(10 - 2x)$ to be defined, the argument must be strictly positive: $10 - 2x > 0$

Solve for $x$ : $-2x > -10$

Dividing by $-2$ reverses the inequality sign: $x < 5$

In interval notation: $x \in (-\infty, 5)$

Find the intersection of both conditions

The domain of $f(x)$ is the intersection of both solution sets: $\text{Domain} = \big((-\infty, -3] \cup [5, \infty)\big) \cap (-\infty, 5)$

For the region $(-\infty, -3]$ , all points satisfy both $x \le -3$ and $x < 5$ , so $(-\infty, -3]$ is included.
For the region $[5, \infty)$ , every value is $\ge 5$ , which conflicts with the strict requirement $x < 5$ . In particular, at $x = 5$ , the logarithmic argument becomes zero, which is undefined. Hence, no points from $[5, \infty)$ belong to the domain.

Therefore, the domain in interval notation is: $(-\infty, -3]$ ](streamdown:incomplete-link)

Graph on the real number line

To represent $(-\infty, -3]$ on a horizontal real number line:

Plot a solid (closed) circle at $x = -3$ to indicate that $-3$ is included in the domain.
Draw a thick shaded line (or ray) extending from $x = -3$ indefinitely to the left, terminating with an arrow pointing toward $-\infty$ .

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student allows the argument of the logarithm to be non-negative instead of strictly positive, incorrectly including x = 5 in the domain.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
x^2 - 2x - 15 $\ge$ 0 $\implies$ (x - 5)(x + 3) $\ge$ 0 $\implies$ x $\le$ -3 $\text{ or }$ x $\ge$ 5
Step 2
10 - 2x $\ge$ 0 $\implies$ -2x $\ge$ -10 $\implies$ x $\le$ 5
Correct
10 - 2x > 0 $\implies$ -2x > -10 $\implies$ x < 5
The argument of a logarithm must be strictly positive ( $> 0$ ), not non-negative ( $\ge 0$ ). At $x = 5$ , the argument is $10 - 2(5) = 0$ , and $\ln(0)$ is undefined.
Step 3
$\text{Intersection: }$ x $\le$ -3 $\text{ or }$ x = 5 $\implies$ (- $\infty$ , -3] $\cup$ $\{$ 5 $\}$
Correct
((- $\infty$ , -3] $\cup$ [5, $\infty$ )) $\cap$ (- $\infty$ , 5) = (- $\infty$ , -3]
Because $x < 5$ , the boundary value $x = 5$ cannot be included in the intersection.
Step 4
$\text{A solid circle at }$ -3 $\text{ with a ray shaded to the left toward }$ - $\infty$$\text{, and a single solid circle at }$ 5.$$
Correct
$\text{A solid closed circle at }$ x = -3 $\text{ with a ray shaded indefinitely to the left toward }$ - $\infty$ .
Since $x = 5$ is not part of the domain, there should not be a point plotted at $x = 5$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly required the radicand of the square root to be non-negative.

You factored the quadratic expression $(x - 5)(x + 3)$ correctly and identified the correct solution intervals for the radical part.

You correctly reversed the inequality direction when dividing by a negative coefficient.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Non-strict inequality for logarithmic argument

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the domain restriction of a natural logarithm as $\ge 0$ instead of strictly $> 0$ , conflating the requirement of an even radical with that of a logarithm.

REASONING PATH

Your ACEpath teacher’s suggested approach

Logarithmic Argument Restriction: For any real logarithm $\ln(g(x))$ , the argument must satisfy $g(x) > 0$ . It can never equal zero or be negative.
Whenever you see $\ln(\dots)$ or $\log_b(\dots)$ , always write a strict inequality ( $> 0$ ) rather than a non-strict inequality ( $\ge 0$ ).

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the logarithmic argument strictly positive: $10 - 2x > 0$ .
Subtract $10$ from both sides: $-2x > -10$ .
Divide by $-2$ and reverse the inequality sign: $x < 5$ .
Find the intersection of $x \in (-\infty, -3] \cup [5, \infty)$ and $x < 5$ : since $[5, \infty)$ does not overlap with $x < 5$ , the only valid interval is $(-\infty, -3]$ .](streamdown:incomplete-link)

Final answer

Domain: $(-\infty, -3]$ ; on the number line, a solid closed circle at $x = -3$ with a shaded ray extending indefinitely to the left.

NEXT STEPS

Final Takeaways from Your Teacher

Memorize the fundamental domain rules side by side: even roots require $\text{radicand} \ge 0$ , whereas logarithms require $\text{argument} > 0$ .
Practice checking endpoint values: test $x = 5$ in the original function to see immediately that $\ln(0)$ is undefined.

OverviewACEpath Teacher's Diagnosis

02 The student forgets to reverse the inequality sign when dividing by negative two, leading to an incorrect domain of x ≥ 5.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
x^2 - 2x - 15 $\ge$ 0 $\implies$ (x - 5)(x + 3) $\ge$ 0 $\implies$ x $\le$ -3 $\text{ or }$ x $\ge$ 5
Step 2
10 - 2x > 0 $\implies$ -2x > -10 $\implies$ x > 5
Correct
10 - 2x > 0 $\implies$ -2x > -10 $\implies$ x < 5
When dividing both sides of $-2x > -10$ by the negative number $-2$ , the inequality sign must flip from $>$ to $<$ , yielding $x < 5$ .
Step 3
$\text{Intersection: } x > 5$
Correct
(- $\infty$ , -3] $\cup$ [5, $\infty$ ) $\cap$ (- $\infty$ , 5) = (- $\infty$ , -3]
Because your linear inequality resulted in $x > 5$ , your intersection calculation followed logically from that error, missing the correct overlap with $(-\infty, -3]$ .
Step 4
(5, $\infty$ ) $\text{ with an open circle at }$ 5 $\text{ and a ray extending rightward}$
Correct
(- $\infty$ , -3] $\text{ with a solid closed circle at }$ -3 $\text{ and a shaded ray extending leftward to }$ - $\infty$
The final interval notation and number line graph followed directly from the incorrect intersection.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly required the radicand to be non-negative and the logarithmic argument to be strictly positive.

You factored the quadratic expression $x^2 - 2x - 15$ correctly and solved for the union of disjoint intervals.

You translated your calculated inequality into consistent interval notation and an accurate number line representation.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Inequality Direction Reversal Error

Dividing both sides of $-2x > -10$ by $-2$ without reversing the inequality sign from $>$ to $<$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You forgot that multiplying or dividing an inequality by a negative quantity reflects numbers across zero, requiring the inequality symbol to reverse direction to maintain truth.

REASONING PATH

Your ACEpath teacher’s suggested approach

Division by Negative Numbers: For any real numbers $a, b$ and $c < 0$ , if $a > b$ , then $\frac{a}{c} < \frac{b}{c}$ .
Whenever you divide or multiply an inequality by a negative constant, immediately flip the inequality symbol.
Testing Boundary Values: A test point on either side of the critical value verifies whether the inequality sign points the correct way.
Substitute $x = 0$ into the original inequality $10 - 2(0) = 10 > 0$ to verify that $x = 0$ must be included in the solution set.

CORRECTED PATH

ACEpath Teacher's Demonstration

Divide both sides by $-2$ and reverse the inequality sign: $x < 5$ .
Intersect the two condition sets: $\big((-\infty, -3] \cup [5, \infty)\big) \cap (-\infty, 5)$ .](streamdown:incomplete-link)
Evaluate the overlap: the region $[5, \infty)$ has no overlap with $x < 5$ , leaving only $(-\infty, -3]$ .

Final answer

Domain in interval notation: $(-\infty, -3]$ ; on the number line, a solid closed circle at $x = -3$ with a shaded ray extending indefinitely to the left.

NEXT STEPS

Final Takeaways from Your Teacher

Practice solving one-step and two-step linear inequalities involving negative coefficients.
Always pick a quick test point such as $x = 0$ to check whether the chosen side of the boundary makes the original inequality true.
Practice graphing overlapping inequalities on stacked number lines before taking their intersection.

OverviewACEpath Teacher's Diagnosis

03 The student takes the union of the two condition sets rather than their intersection, resulting in an overly broad domain of all real numbers.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$x^2 - 2x - 15 \ge 0 \implies (x - 5)(x + 3) \ge 0 \implies x \le -3 \text{ or } x \ge 5$
Step 2
$10 - 2x > 0 \implies -2x > -10 \implies x < 5$
Step 3
$\text{Domain} = (x \le -3 \text{ or } x \ge 5) \cup (x < 5) = \text{all real numbers}$
Correct
$\text{Domain} = ((-\infty, -3] \cup [5, \infty)) \cap (-\infty, 5) = (-\infty, -3]$
You combined the conditions with a union ( $\cup$ ) instead of an intersection ( $\cap$ ). A sum of two functions requires both terms to be defined at the same time, meaning you must find the values of $x$ satisfying both conditions simultaneously.
Step 4
Interval: $(-\infty, \infty)$ ; Graph: entire number line shaded in both directions
Correct
Interval: $(-\infty, -3]$ ; Graph: a solid closed circle at $x = -3$ with a shaded ray extending indefinitely to the left toward $-\infty$
Because the union was computed in the previous step, the resulting interval and graphed number line erroneously included all real numbers.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly established the non-negativity condition for the radicand and solved the quadratic inequality.

You correctly set the logarithmic argument to be strictly positive and accurately reversed the inequality when dividing by a negative number.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Set Operation Confusion (Union vs. Intersection)

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated simultaneous function domain requirements as an 'either/or' scenario (union) rather than an 'and' scenario (intersection). When combining multiple expressions in a single function, the overall domain is strictly the intersection of their individual domains.

REASONING PATH

Your ACEpath teacher’s suggested approach

Simultaneous Function Domains: For a function $f(x) = g(x) + h(x)$ , both $g(x)$ and $h(x)$ must be defined at the same input value $x$ .
Always take the intersection ( $\text{Domain}(g) \cap \text{Domain}(h)$ ), not the union.
Testing Values in Candidate Intervals: If you suspect an interval belongs to the domain, plugging a test point into the original function will confirm if any component fails.
Test a point like $x = 0$ : $\sqrt{0^2 - 2(0) - 15} = \sqrt{-15}$ , which is not real, confirming $x = 0$ cannot be in the domain.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set up the domain intersection: $\text{Domain} = ((-\infty, -3] \cup [5, \infty)) \cap (-\infty, 5)$ .](streamdown:incomplete-link)
Find the common overlapping values: $(-\infty, -3]$ overlaps with $(-\infty, 5)$ , while $[5, \infty)$ has no overlap with $(-\infty, 5)$ because $x = 5$ is excluded by the strict inequality $x < 5$ .](streamdown:incomplete-link)
Write the final domain in interval notation: $(-\infty, -3]$ .
Graph this interval on a real number line: place a solid closed circle at $x = -3$ and draw a shaded ray extending to the left with an arrow pointing toward $-\infty$ .

Final answer

Domain: $(-\infty, -3]$ ; Graph: a solid closed circle at $x = -3$ with a shaded ray extending indefinitely to the left.

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between 'AND' (intersection $\cap$ ) for simultaneous domain constraints versus 'OR' (union $\cup$ ) for piecewise branches.
Practice drawing two inequalities on separate parallel number lines before reading off their common overlapping region.
Check candidate intervals by substituting test values into all original terms of the function.

Lock it in

Question

What condition must an expression $g(x)$ satisfy for $\ln(g(x))$ to be defined over the real numbers?

Answer

$g(x) > 0$ . The argument of a logarithm must be strictly positive because $\ln(0)$ and $\ln(\text{negative})$ are undefined.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student allows the argument of the logarithm to be non-negative instead of strictly positive, incorrectly including x = 5 in the domain.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student forgets to reverse the inequality sign when dividing by negative two, leading to an incorrect domain of x ≥ 5.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student takes the union of the two condition sets rather than their intersection, resulting in an overly broad domain of all real numbers.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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