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Question 1
Consider the function $q (x) = x^{6} - 5 x^{4} + 3 x^{2} - 7$ . Which of the following correctly classifies $q$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BEven, because $q (- x) = q (x)$ for all $x$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: State the condition for an even function
A function is even if $q (- x) = q (x)$ for all $x$ in its domain.
Step 2: Compute $q(-x)$
$q (- x) = (- x)^{6} - 5 (- x)^{4} + 3 (- x)^{2} - 7 = x^{6} - 5 x^{4} + 3 x^{2} - 7$
Step 3: Compare with $q(x)$
We see $q (- x) = x^{6} - 5 x^{4} + 3 x^{2} - 7 = q (x)$ . Every term has an even power (including the constant, degree 0), so the function is even.
Step 4: State conclusion
Since $q (- x) = q (x)$ , $q$ is even. The graph is symmetric about the $y$ -axis.
Method #2Approach 2
Step 1: Identify what is being tested
We need to classify the function as odd, even, or neither by checking the structure of the polynomial.
Step 2: Eliminate 'Odd, because all exponents are even'
A polynomial with all even-power terms satisfies $q (- x) = q (x)$ , not $q (- x) = - q (x)$ . Even-power terms indicate an even function, not odd. This option contains a factual contradiction.
Step 3: Eliminate 'Neither odd nor even'
The claim references a constant term causing issues. The constant $- 7$ is degree 0, which is an even degree, so it does not break the even symmetry. A constant term only prevents odd classification, which is irrelevant here.
Step 4: Eliminate 'Even, because the graph passes through the origin'
Passing through the origin is not a criterion for being even. In fact, $q (0) = - 7 \neq = 0$ , so the graph does not even pass through the origin. This reason is false.
Step 5: Select the correct answer
The correct answer is 'Even, because $q (- x) = q (x)$ for all $x$ ' — this is the definition of an even function, verified by the calculation above.
Question 2
Let $f (x) = x 9 - x^{2}$ for $- 3 \leq x \leq 3$ . Which statement correctly classifies $f$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BOdd, because $f (- x) = - f (x)$ for all $x$ in $[- 3, 3]$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the classification condition
A function is odd if $f (- x) = - f (x)$ for all $x$ in its domain. The domain $[- 3, 3]$ is symmetric about $0$ , which is a necessary condition.
Step 2: Compute $f(-x)$
$f (- x) = (- x) 9 - (- x)^{2} = (- x) 9 - x^{2} = - x 9 - x^{2}$
Step 3: Compare with $-f(x)$
$- f (x) = - x 9 - x^{2}$ Since $f (- x) = - x 9 - x^{2} = - f (x)$ , the odd condition is satisfied.
Step 4: Conclude
The function is odd. Its graph has 180° rotational symmetry about the origin.
Method #2Approach 2
Step 1: Identify what is being tested
We must compute $f (- x)$ and compare it to both $f (x)$ and $- f (x)$ .
Step 2: Eliminate 'Even, because the domain is symmetric about 0'
A symmetric domain is a necessary condition for even or odd classification, but not sufficient. We must verify $f (- x) = f (x)$ : since $f (- x) = - x 9 - x^{2} \neq = x 9 - x^{2} = f (x)$ (for $x \neq = 0$ ), the function is not even.
Step 3: Eliminate 'Neither, because...'
The reasoning is logically flawed — a product of two functions can certainly be odd (e.g. $x \cdot cos x$ is odd). This option is eliminated on conceptual grounds.
Step 4: Eliminate 'Odd, because the domain includes negative values'
Including negative values in the domain is again only a necessary condition, not a reason for being odd. The correct reason must invoke the algebraic identity $f (- x) = - f (x)$ .
Step 5: Select the correct answer
'Odd, because $f (- x) = - f (x)$ for all $x$ in $[- 3, 3]$ ' is correct, as confirmed by the direct computation.

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13 more questions in this topic

Free preview 2/15

Question 1
Consider the function $q (x) = x^{6} - 5 x^{4} + 3 x^{2} - 7$ . Which of the following correctly classifies $q$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BEven, because $q (- x) = q (x)$ for all $x$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: State the condition for an even function
A function is even if $q (- x) = q (x)$ for all $x$ in its domain.
Step 2: Compute $q(-x)$
$q (- x) = (- x)^{6} - 5 (- x)^{4} + 3 (- x)^{2} - 7 = x^{6} - 5 x^{4} + 3 x^{2} - 7$
Step 3: Compare with $q(x)$
We see $q (- x) = x^{6} - 5 x^{4} + 3 x^{2} - 7 = q (x)$ . Every term has an even power (including the constant, degree 0), so the function is even.
Step 4: State conclusion
Since $q (- x) = q (x)$ , $q$ is even. The graph is symmetric about the $y$ -axis.
Method #2Approach 2
Step 1: Identify what is being tested
We need to classify the function as odd, even, or neither by checking the structure of the polynomial.
Step 2: Eliminate 'Odd, because all exponents are even'
A polynomial with all even-power terms satisfies $q (- x) = q (x)$ , not $q (- x) = - q (x)$ . Even-power terms indicate an even function, not odd. This option contains a factual contradiction.
Step 3: Eliminate 'Neither odd nor even'
The claim references a constant term causing issues. The constant $- 7$ is degree 0, which is an even degree, so it does not break the even symmetry. A constant term only prevents odd classification, which is irrelevant here.
Step 4: Eliminate 'Even, because the graph passes through the origin'
Passing through the origin is not a criterion for being even. In fact, $q (0) = - 7 \neq = 0$ , so the graph does not even pass through the origin. This reason is false.
Step 5: Select the correct answer
The correct answer is 'Even, because $q (- x) = q (x)$ for all $x$ ' — this is the definition of an even function, verified by the calculation above.
Question 2
Let $f (x) = x 9 - x^{2}$ for $- 3 \leq x \leq 3$ . Which statement correctly classifies $f$ ?
No clue? Show me the answer
Correct answerCorrect!Incorrect
BOdd, because $f (- x) = - f (x)$ for all $x$ in $[- 3, 3]$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the classification condition
A function is odd if $f (- x) = - f (x)$ for all $x$ in its domain. The domain $[- 3, 3]$ is symmetric about $0$ , which is a necessary condition.
Step 2: Compute $f(-x)$
$f (- x) = (- x) 9 - (- x)^{2} = (- x) 9 - x^{2} = - x 9 - x^{2}$
Step 3: Compare with $-f(x)$
$- f (x) = - x 9 - x^{2}$ Since $f (- x) = - x 9 - x^{2} = - f (x)$ , the odd condition is satisfied.
Step 4: Conclude
The function is odd. Its graph has 180° rotational symmetry about the origin.
Method #2Approach 2
Step 1: Identify what is being tested
We must compute $f (- x)$ and compare it to both $f (x)$ and $- f (x)$ .
Step 2: Eliminate 'Even, because the domain is symmetric about 0'
A symmetric domain is a necessary condition for even or odd classification, but not sufficient. We must verify $f (- x) = f (x)$ : since $f (- x) = - x 9 - x^{2} \neq = x 9 - x^{2} = f (x)$ (for $x \neq = 0$ ), the function is not even.
Step 3: Eliminate 'Neither, because...'
The reasoning is logically flawed — a product of two functions can certainly be odd (e.g. $x \cdot cos x$ is odd). This option is eliminated on conceptual grounds.
Step 4: Eliminate 'Odd, because the domain includes negative values'
Including negative values in the domain is again only a necessary condition, not a reason for being odd. The correct reason must invoke the algebraic identity $f (- x) = - f (x)$ .
Step 5: Select the correct answer
'Odd, because $f (- x) = - f (x)$ for all $x$ in $[- 3, 3]$ ' is correct, as confirmed by the direct computation.

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Step-by-step walkthrough

Step 1: State the condition for an even function

Step 2: Compute $q(-x)$

Step 3: Compare with $q(x)$

Step 4: State conclusion

Step 1: Identify what is being tested

Step 2: Eliminate 'Odd, because all exponents are even'

Step 3: Eliminate 'Neither odd nor even'

Step 4: Eliminate 'Even, because the graph passes through the origin'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Identify the classification condition

Step 2: Compute $f(-x)$

Step 3: Compare with $-f(x)$

Step 4: Conclude

Step 1: Identify what is being tested

Step 2: Eliminate 'Even, because the domain is symmetric about 0'

Step 3: Eliminate 'Neither, because...'

Step 4: Eliminate 'Odd, because the domain includes negative values'

Step 5: Select the correct answer

13 more questions in this topic

Step-by-step walkthrough

Step 1: State the condition for an even function

Step 2: Compute $q(-x)$

Step 3: Compare with $q(x)$

Step 4: State conclusion

Step 1: Identify what is being tested

Step 2: Eliminate 'Odd, because all exponents are even'

Step 3: Eliminate 'Neither odd nor even'

Step 4: Eliminate 'Even, because the graph passes through the origin'

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Identify the classification condition

Step 2: Compute $f(-x)$

Step 3: Compare with $-f(x)$

Step 4: Conclude

Step 1: Identify what is being tested

Step 2: Eliminate 'Even, because the domain is symmetric about 0'

Step 3: Eliminate 'Neither, because...'

Step 4: Eliminate 'Odd, because the domain includes negative values'

Step 5: Select the correct answer

13 more questions in this topic