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Sagot :
To determine whether the function [tex]\( f(x) = 2x^3 - 6 \)[/tex] is one-to-one, we can use the derivative test, which involves checking the behavior of the function's derivative.
1. Find the derivative of the function:
The derivative of [tex]\( f(x) = 2x^3 - 6 \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 6) = 6x^2. \][/tex]
2. Identify critical points:
Critical points occur where the derivative is zero. Set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 6x^2 = 0 \Rightarrow x^2 = 0 \Rightarrow x = 0. \][/tex]
So, the critical point is [tex]\( x = 0 \)[/tex].
3. Determine the nature of critical points:
Next, we check whether the derivative changes sign around the critical point. Evaluate the second derivative to check the concavity at [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}(6x^2) = 12x. \][/tex]
At [tex]\( x = 0 \)[/tex], the second derivative [tex]\( f''(0) = 12 \times 0 = 0 \)[/tex], which does not tell us about concavity. However, for [tex]\( x < 0 \)[/tex] and [tex]\( x > 0 \)[/tex], the derivative [tex]\( 6x^2 \)[/tex] remains non-negative.
4. Test the behavior of [tex]\( f'(x) \)[/tex]:
Since the derivative [tex]\( 6x^2 \geq 0 \)[/tex] for all [tex]\( x \)[/tex] and only touches zero at [tex]\( x = 0 \)[/tex], this implies that the function is non-decreasing at [tex]\( x = 0 \)[/tex]. Outside of this point, [tex]\( 6x^2 > 0 \)[/tex], indicating the function is strictly increasing or decreasing in any interval that does not include [tex]\( x = 0 \)[/tex].
5. Conclusion:
Because the function is monotonic and does not change direction (it either increases or remains flat, but does not decrease), [tex]\( f(x) = 2x^3 - 6 \)[/tex] is one-to-one. For every [tex]\( x \)[/tex]-value, there is a unique corresponding [tex]\( f(x) \)[/tex]-value, and vice versa.
Thus, the correct choice 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.
1. Find the derivative of the function:
The derivative of [tex]\( f(x) = 2x^3 - 6 \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 6) = 6x^2. \][/tex]
2. Identify critical points:
Critical points occur where the derivative is zero. Set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 6x^2 = 0 \Rightarrow x^2 = 0 \Rightarrow x = 0. \][/tex]
So, the critical point is [tex]\( x = 0 \)[/tex].
3. Determine the nature of critical points:
Next, we check whether the derivative changes sign around the critical point. Evaluate the second derivative to check the concavity at [tex]\( x = 0 \)[/tex]:
[tex]\[ f''(x) = \frac{d}{dx}(6x^2) = 12x. \][/tex]
At [tex]\( x = 0 \)[/tex], the second derivative [tex]\( f''(0) = 12 \times 0 = 0 \)[/tex], which does not tell us about concavity. However, for [tex]\( x < 0 \)[/tex] and [tex]\( x > 0 \)[/tex], the derivative [tex]\( 6x^2 \)[/tex] remains non-negative.
4. Test the behavior of [tex]\( f'(x) \)[/tex]:
Since the derivative [tex]\( 6x^2 \geq 0 \)[/tex] for all [tex]\( x \)[/tex] and only touches zero at [tex]\( x = 0 \)[/tex], this implies that the function is non-decreasing at [tex]\( x = 0 \)[/tex]. Outside of this point, [tex]\( 6x^2 > 0 \)[/tex], indicating the function is strictly increasing or decreasing in any interval that does not include [tex]\( x = 0 \)[/tex].
5. Conclusion:
Because the function is monotonic and does not change direction (it either increases or remains flat, but does not decrease), [tex]\( f(x) = 2x^3 - 6 \)[/tex] is one-to-one. For every [tex]\( x \)[/tex]-value, there is a unique corresponding [tex]\( f(x) \)[/tex]-value, and vice versa.
Thus, the correct choice 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.
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