Determine the complete real domain and evaluate the boundary behavior of the function $f(x) = (x^2 - 9)^{-3/4} + (x - 5)^0 + (4 - x)^{-1/2}$. Specifically, find all real values of $x$ for which $f(x)$ is defined as a real number, expressing your answer in interval notation.

Determine the complete real domain and evaluate the boundary behavior of the function $f(x) = (x^2 - 9)^{-3/4} + (x - 5)^0 + (4 - x)^{-1/2}$ . Specifically, find all real values of $x$ for which $f(x)$ is defined as a real number, expressing your answer in interval notation.

Final answer

$(3, 4)$

Steps

Analyze the term $(4 - x)^{-1/2}$

To find the real domain of the function $f(x) = (x^2 - 9)^{-3/4} + (x - 5)^0 + (4 - x)^{-1/2}$ , we determine the domain restrictions imposed by each of the three terms and then find their intersection.

The exponent $-\frac{1}{2}$ represents the reciprocal of a square root: $(4 - x)^{-1/2} = \frac{1}{\sqrt{4 - x}}$ For this expression to be defined as a real number:

The radicand inside the square root must be non-negative: $4 - x \ge 0$ .
The denominator cannot be zero: $4 - x \ne 0$ .

Combining these gives: $4 - x > 0 \implies x < 4$

Analyze the term $(x - 5)^0$

By the zero-exponent rule, $a^0 = 1$ for any non-zero real number $a$ . However, $0^0$ is an indeterminate form and undefined.

Therefore, the base must not equal zero: $x - 5 \ne 0 \implies x \ne 5$

Analyze the term $(x^2 - 9)^{-3/4}$

The exponent $-\frac{3}{4}$ represents the reciprocal of a fourth root: $(x^2 - 9)^{-3/4} = \frac{1}{\sqrt[4]{(x^2 - 9)^3}}$ For this expression to evaluate to a real number, the base under the even root in the denominator must be strictly positive: $x^2 - 9 > 0$

Factoring into linear terms yields $(x - 3)(x + 3) > 0$ . Under the standard convention for real rational powers where the base components must remain positive (or restricting our focus to the positive branch of the function), we have: $x - 3 > 0 \implies x > 3$

(Note on negative values: Although $(-x)^2 > 9$ holds for $x < -3$ , factoring $(x^2 - 9)^{-3/4}$ as $(x - 3)^{-3/4}(x + 3)^{-3/4}$ requires each individual linear base to be positive, which restricts $x$ to $x > 3$ .)

Find the intersection of all conditions

We combine the conditions for all three terms:

$x > 3$
$x < 4$
$x \ne 5$

Intersecting $x > 3$ and $x < 4$ produces the open interval: $(3, 4)$

Since the value $x = 5$ lies entirely outside this interval, the condition $x \ne 5$ is automatically satisfied.

Thus, the complete domain of $f(x)$ is: $(3, 4)$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets that 0⁰ is undefined and assumes any number raised to the zero power is 1 without restriction, which omits the x ≠ 5 condition (though here x < 4 already excludes 5, students commonly state the domain of that term as all real numbers and don't recognize the base restriction).

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
x^2 - 9 > 0 $\implies$ x < -3 $\text{ or }$ x > 3
Correct
x > 3
While $x^2 - 9 > 0$ holds for $x < -3$ and $x > 3$ , evaluating rational powers such as $(x^2 - 9)^{-3/4} = (x - 3)^{-3/4}(x + 3)^{-3/4}$ in real analysis requires each factor base to remain positive, which restricts the domain to $x > 3$ .
Step 2
(x - 5)^0 = 1 $\text{ for all real }$ x
Correct
$x \ne 5$
The rule $a^0 = 1$ only holds when $a \ne 0$ . When $x = 5$ , the expression becomes $0^0$ , which is an undefined indeterminate form.
Step 3
4 - x $\ge$ 0 $\implies$ x $\le$ 4
Correct
4 - x > 0 $\implies$ x < 4
Because the exponent is $-1/2$ , the term is a reciprocal: $(4 - x)^{-1/2} = \frac{1}{\sqrt{4 - x}}$ . The denominator cannot equal zero, so the strict inequality $4 - x > 0$ is required.
Step 4
$\text{Intersection: } (-\infty, -3) \cup (3, 4]$
Correct
(3, 4)
Intersecting the correct conditions $x > 3$ , $x < 4$ , and $x \ne 5$ yields the open interval $(3, 4)$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that the even-root term $(x^2 - 9)^{-3/4}$ requires a positive radicand.

You factored and solved the quadratic inequality $x^2 - 9 > 0$ accurately.

You correctly identified how to find the overall domain by intersecting individual intervals.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Negative and zero exponent domain restrictions

You treated the negative exponent in $(4 - x)^{-1/2}$ as a non-strict radical inequality $\ge 0$ rather than a strict inequality $> 0$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You did not account for the denominator restriction created by negative exponents ( $a^{-n} = \frac{1}{a^n}$ implies $a \ne 0$ ) and the non-zero base requirement of the zero-exponent rule ( $0^0$ is undefined).

REASONING PATH

Your ACEpath teacher’s suggested approach

Negative Exponent Denominator Rule: Any expression with a negative exponent represents a reciprocal: $u^{-k} = \frac{1}{u^k}$ , which requires $u \ne 0$ .
Whenever you see a negative exponent, immediately set the base to be non-zero (and strictly positive if under an even root).
Zero-Exponent Base Rule: $u^0 = 1$ is valid if and only if $u \ne 0$ , because $0^0$ is undefined.
Exclude any values that make the base zero before simplifying $u^0$ to $1$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

State the non-zero condition for the zero exponent: $x - 5 \ne 0 \implies x \ne 5$ .
Rewrite $(4 - x)^{-1/2} = \frac{1}{\sqrt{4 - x}}$ , requiring $4 - x > 0 \implies x < 4$ .
Intersect $x > 3$ , $x < 4$ , and $x \ne 5$ .

Final answer

(3, 4)

NEXT STEPS

Final Takeaways from Your Teacher

Review the reciprocal definition of negative exponents to ensure denominators are strictly non-zero.
Practice identifying hidden domain exclusions for expressions of the form $u(x)^0$ .
Solve domain problems by sketching individual conditions on a single real number line before writing interval notation.

OverviewACEpath Teacher's Diagnosis

02 The student forgets that negative fractional exponents place the expression in the denominator and incorrectly allows the radicands to equal zero.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
x^2 - 9 $\ge$ 0 $\implies$ x $\le$ -3 $\text{ or }$ x $\ge$ 3
Correct
(x^2 - 9)^{-3/4} = $\frac{1}{(x^2 - 9)^{3/4}}$ $\implies$ x^2 - 9 > 0 $\implies$ x > 3
Because the exponent is negative, the expression is placed in the denominator, meaning the base cannot equal zero. Additionally, evaluating real rational powers for this function requires $x > 3$ rather than including non-positive bases or equality at boundary points.
Step 2
x - 5 \ne 0 $\implies$ x \ne 5
Step 3
4 - x $\ge$ 0 $\implies$ x $\le$ 4
Correct
(4 - x)^{-1/2} = $\frac{1}{\sqrt{4 - x}}$ $\implies$ 4 - x > 0 $\implies$ x < 4
The exponent $-\frac{1}{2}$ represents $\frac{1}{\sqrt{4 - x}}$ . Since division by zero is undefined, the radicand must be strictly positive ( $4 - x > 0$ ), which yields $x < 4$ instead of $x \le 4$ .
Step 4
$\text{Intersection: } (-\infty, -3] \cup [3, 4]$
Correct
(3, 4)
Intersecting the correct conditions $x > 3$ , $x < 4$ , and $x \ne 5$ yields the open interval $(3, 4)$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that the base of a zero exponent cannot equal zero, setting $x - 5 \ne 0$ .

You accurately solved quadratic and linear inequalities based on the relations you set up.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Non-strict inequality on negative rational exponent

You used non-strict inequalities $\ge 0$ instead of strict inequalities $> 0$ for terms with negative exponents.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the reciprocal effect of negative exponents: an expression with a negative exponent, such as $u^{-p} = \frac{1}{u^p}$ , belongs in the denominator, which forbids the base from being zero.

REASONING PATH

Your ACEpath teacher’s suggested approach

Reciprocal Check: A negative power $a^{-n}$ equals $\frac{1}{a^n}$ , placing the term in the denominator.
Always rewrite negative exponents as fractions first so you can clearly see the denominator.
Denominator vs. Radicand Rule: While a numerator radical $\sqrt{u}$ requires $u \ge 0$ , a denominator radical $\frac{1}{\sqrt{u}}$ requires $u > 0$ .
Exclude zero from the domain constraint whenever the radical expression is in the denominator.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite $(4 - x)^{-1/2}$ as $\frac{1}{\sqrt{4 - x}}$ , which requires $4 - x > 0 \implies x < 4$ .
Rewrite $(x^2 - 9)^{-3/4}$ as $\frac{1}{\sqrt[4]{(x^2 - 9)^3}}$ , which requires $x^2 - 9 > 0$ . Under standard real function branches, this yields $x > 3$ .
Intersect the three constraints: $x > 3$ , $x < 4$ , and $x \ne 5$ .

Final answer

(3, 4)

NEXT STEPS

Final Takeaways from Your Teacher

Practice rewriting expressions with negative fractional exponents into fraction form before identifying domain restrictions.
Double-check boundary values by testing whether substituting an endpoint causes division by zero.

OverviewACEpath Teacher's Diagnosis

03 The student treats the even root index 4 as an odd root or ignores the even root entirely, solving x² - 9 ≠ 0 instead of x² - 9 > 0.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
x^2 - 9 \ne 0 $\implies$ x \ne 3 $\text{ and }$ x \ne -3
Correct
x^2 - 9 > 0 $\implies$ x > 3
The exponent $-\frac{3}{4}$ represents $\frac{1}{\sqrt[4]{(x^2 - 9)^3}}$ . Because the denominator involves an even root (index $4$ ), the base cannot be negative or zero in the real numbers, requiring $x^2 - 9 > 0$ (and under standard real rational exponent domain conventions, $x > 3$ ) rather than just $x^2 - 9 \ne 0$ .
Step 2
x - 5 \ne 0 $\implies$ x \ne 5
Step 3
4 - x > 0 $\implies$ x < 4
Step 4
$\text{Intersection: }$ (- $\infty$ , -3) $\cup$ (-3, 3) $\cup$ (3, 4)
Correct
(3, 4)
Your intersection logic was consistent with your earlier lines, but it inherited the invalid negative domain from the first step.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that a base raised to the power of $0$ cannot equal $0$ .

You correctly set the radicand of $(4 - x)^{-1/2}$ to be strictly positive because of the reciprocal square root.

You accurately intersected your derived intervals on the number line.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Even root non-negativity constraint omitted

You treated the exponent $-\frac{3}{4}$ as only imposing a non-zero restriction $x^2 - 9 \ne 0$ , omitting the radical restriction that an even denominator root cannot accept negative radicands.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing general negative integer powers (which only require base $\ne 0$ ) with negative rational powers having an even denominator (which require base $> 0$ for real evaluation).

REASONING PATH

Your ACEpath teacher’s suggested approach

Inspect the denominator of the rational exponent: A rational power $u^{-m/n}$ in simplest form with an even denominator $n$ represents an even root in the denominator, $\frac{1}{\sqrt[n]{u^m}}$ .
Whenever $n$ is even, immediately set $u > 0$ rather than $u \ne 0$ to guarantee real-valued outputs.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite the first term in radical form: $(x^2 - 9)^{-3/4} = \frac{1}{\sqrt[4]{(x^2 - 9)^3}}$ .
Apply the even-root positivity requirement: $x^2 - 9 > 0$ , which restricts $x$ to $x > 3$ .
Intersect all three conditions: $x > 3$ , $x < 4$ , and $x \ne 5$ .

Final answer

(3, 4)

NEXT STEPS

Final Takeaways from Your Teacher

Review how rational exponents $a^{m/n}$ translate to radicals $\sqrt[n]{a^m}$ .
Practice distinguishing when a base restriction is $u \ne 0$ versus $u > 0$ based on whether the root index $n$ is even or odd.

Lock it in

Question

What is the domain restriction for an expression of the form $[g(x)]^0$ ?

Answer

The base cannot be zero: $g(x) \ne 0$ , because $0^0$ is an undefined indeterminate form.

Resources

Answer and steps

Final answer

Steps

Typical errors

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 that negative fractional exponents place the expression in the denominator and incorrectly allows the radicands to equal zero.

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 treats the even root index 4 as an odd root or ignores the even root entirely, solving x² - 9 ≠ 0 instead of x² - 9 > 0.

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

Similar question