Answer :
To determine whether the function [tex]\( f(x) = 2x^3 - 5 \)[/tex] is one-to-one, we need to verify if each [tex]\( x \)[/tex]-value corresponds to exactly one [tex]\( f(x) \)[/tex]-value, and vice versa.
Here's a step-by-step verification:
1. Definition of One-to-One Function: A function [tex]\( f \)[/tex] is one-to-one if for every [tex]\( a \)[/tex] and [tex]\( b \)[/tex] in the domain of [tex]\( f \)[/tex], whenever [tex]\( f(a) = f(b) \)[/tex], it must be that [tex]\( a = b \)[/tex].
2. Derivative Analysis: A function is often one-to-one if it is either strictly increasing or strictly decreasing. We can determine this by examining the derivative of the function.
3. Find the Derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 5) = 6x^2 \][/tex]
4. Check the Critical Points: Set the derivative equal to zero to find any critical points.
[tex]\[ 6x^2 = 0 \implies x = 0 \][/tex]
5. Analyze the Derivative: The derivative [tex]\( f'(x) = 6x^2 \)[/tex] is zero only at [tex]\( x = 0 \)[/tex].
- When [tex]\( x < 0 \)[/tex], [tex]\( 6x^2 > 0 \)[/tex].
- When [tex]\( x > 0 \)[/tex], [tex]\( 6x^2 > 0 \)[/tex].
Essentially, [tex]\( f'(x) = 6x^2 \)[/tex] is always non-negative and only zero at a single point ( [tex]\( x = 0 \)[/tex] ). This indicates that [tex]\( f(x) \)[/tex] is non-decreasing. However, [tex]\( 6x^2 > 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex], meaning the function is strictly increasing everywhere except at [tex]\( x = 0 \)[/tex].
6. Conclusion: Since the function [tex]\( f(x) = 2x^3 - 5 \)[/tex] is strictly increasing (as [tex]\( f'(x) > 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex]), every horizontal line will intersect the graph of the function at exactly one point. This satisfies the horizontal line test, ensuring each [tex]\( f(x) \)[/tex]-value corresponds to exactly one [tex]\( x \)[/tex]-value and vice versa.
Therefore, the correct answer is:
B. Yes, because each [tex]\( x \)[/tex]-value corresponds to only one [tex]\( f(x) \)[/tex]-value, and each [tex]\( f(x) \)[/tex]-value corresponds to only one [tex]\( x \)[/tex]-value.
Here's a step-by-step verification:
1. Definition of One-to-One Function: A function [tex]\( f \)[/tex] is one-to-one if for every [tex]\( a \)[/tex] and [tex]\( b \)[/tex] in the domain of [tex]\( f \)[/tex], whenever [tex]\( f(a) = f(b) \)[/tex], it must be that [tex]\( a = b \)[/tex].
2. Derivative Analysis: A function is often one-to-one if it is either strictly increasing or strictly decreasing. We can determine this by examining the derivative of the function.
3. Find the Derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 5) = 6x^2 \][/tex]
4. Check the Critical Points: Set the derivative equal to zero to find any critical points.
[tex]\[ 6x^2 = 0 \implies x = 0 \][/tex]
5. Analyze the Derivative: The derivative [tex]\( f'(x) = 6x^2 \)[/tex] is zero only at [tex]\( x = 0 \)[/tex].
- When [tex]\( x < 0 \)[/tex], [tex]\( 6x^2 > 0 \)[/tex].
- When [tex]\( x > 0 \)[/tex], [tex]\( 6x^2 > 0 \)[/tex].
Essentially, [tex]\( f'(x) = 6x^2 \)[/tex] is always non-negative and only zero at a single point ( [tex]\( x = 0 \)[/tex] ). This indicates that [tex]\( f(x) \)[/tex] is non-decreasing. However, [tex]\( 6x^2 > 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex], meaning the function is strictly increasing everywhere except at [tex]\( x = 0 \)[/tex].
6. Conclusion: Since the function [tex]\( f(x) = 2x^3 - 5 \)[/tex] is strictly increasing (as [tex]\( f'(x) > 0 \)[/tex] for all [tex]\( x \neq 0 \)[/tex]), every horizontal line will intersect the graph of the function at exactly one point. This satisfies the horizontal line test, ensuring each [tex]\( f(x) \)[/tex]-value corresponds to exactly one [tex]\( x \)[/tex]-value and vice versa.
Therefore, the correct answer is:
B. Yes, because each [tex]\( x \)[/tex]-value corresponds to only one [tex]\( f(x) \)[/tex]-value, and each [tex]\( f(x) \)[/tex]-value corresponds to only one [tex]\( x \)[/tex]-value.
