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Sagot :
To determine which statement best describes the function [tex]\( f(x) = x^3 - x^2 - 9x + 9 \)[/tex], we need to analyze the function step by step.
### Step 1: Verify if it is a function
A function is a relation where each input (x-value) corresponds to exactly one output (y-value). The given expression [tex]\( f(x) = x^3 - x^2 - 9x + 9 \)[/tex] is a polynomial of degree 3, which inherently fits the definition of a function since for every [tex]\( x \)[/tex] there is a unique resulting [tex]\( y \)[/tex]. Therefore, it is indeed a function.
### Step 2: Vertical Line Test
The vertical line test helps determine if a graph represents a function. If any vertical line intersects the graph at more than one point, the graph does not represent a function. Since [tex]\( f(x) \)[/tex] is a polynomial, which is continuous and smooth, it satisfies the vertical line test. Thus, it is unquestionably a function.
### Step 3: Determine One-to-One or Many-to-One
To check if the function is one-to-one or many-to-one, we inspect its behavior. A function is one-to-one if different inputs produce different outputs, which implies that [tex]\( f(x_1) = f(x_2) \)[/tex] only when [tex]\( x_1 = x_2 \)[/tex]. For this, we need to investigate the derivative of the function.
First, find the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 3x^2 - 2x - 9 \][/tex]
We need to check the critical points by setting the first derivative equal to zero:
[tex]\[ 3x^2 - 2x - 9 = 0 \][/tex]
Solve the quadratic equation:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-9)}}{2(3)} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 108}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{112}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm 4\sqrt{7}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm 2\sqrt{7}}{3} \][/tex]
There are two critical points, which we denote as [tex]\( x_1 = \frac{1 + 2\sqrt{7}}{3} \)[/tex] and [tex]\( x_2 = \frac{1 - 2\sqrt{7}}{3} \)[/tex].
Now, check the nature of these critical points by evaluating the second derivative:
[tex]\[ f''(x) = 6x - 2 \][/tex]
Evaluate the second derivative at the critical points:
[tex]\[ f''\left(\frac{1 + 2\sqrt{7}}{3}\right) = 6\left(\frac{1 + 2\sqrt{7}}{3}\right) - 2 = 2 + 4\sqrt{7} > 0 \][/tex]
[tex]\[ f''\left(\frac{1 - 2\sqrt{7}}{3}\right) = 6\left(\frac{1 - 2\sqrt{7}}{3}\right) - 2 = 2 - 4\sqrt{7} < 0 \][/tex]
Since the second derivative changes sign at these critical points, [tex]\( f(x) \)[/tex] has both a local minimum and a local maximum. Therefore, the function is not one-to-one because it does not maintain a consistent increasing or decreasing trend across all values of [tex]\( x \)[/tex].
### Conclusion
Since [tex]\( f(x) \)[/tex] passes the vertical line test and has multiple y-values for some x-values (due to local maxima and minima), the function is many-to-one.
The correct answer is:
C. It is a many-to-one function.
### Step 1: Verify if it is a function
A function is a relation where each input (x-value) corresponds to exactly one output (y-value). The given expression [tex]\( f(x) = x^3 - x^2 - 9x + 9 \)[/tex] is a polynomial of degree 3, which inherently fits the definition of a function since for every [tex]\( x \)[/tex] there is a unique resulting [tex]\( y \)[/tex]. Therefore, it is indeed a function.
### Step 2: Vertical Line Test
The vertical line test helps determine if a graph represents a function. If any vertical line intersects the graph at more than one point, the graph does not represent a function. Since [tex]\( f(x) \)[/tex] is a polynomial, which is continuous and smooth, it satisfies the vertical line test. Thus, it is unquestionably a function.
### Step 3: Determine One-to-One or Many-to-One
To check if the function is one-to-one or many-to-one, we inspect its behavior. A function is one-to-one if different inputs produce different outputs, which implies that [tex]\( f(x_1) = f(x_2) \)[/tex] only when [tex]\( x_1 = x_2 \)[/tex]. For this, we need to investigate the derivative of the function.
First, find the first derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = 3x^2 - 2x - 9 \][/tex]
We need to check the critical points by setting the first derivative equal to zero:
[tex]\[ 3x^2 - 2x - 9 = 0 \][/tex]
Solve the quadratic equation:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-9)}}{2(3)} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 108}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{112}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm 4\sqrt{7}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm 2\sqrt{7}}{3} \][/tex]
There are two critical points, which we denote as [tex]\( x_1 = \frac{1 + 2\sqrt{7}}{3} \)[/tex] and [tex]\( x_2 = \frac{1 - 2\sqrt{7}}{3} \)[/tex].
Now, check the nature of these critical points by evaluating the second derivative:
[tex]\[ f''(x) = 6x - 2 \][/tex]
Evaluate the second derivative at the critical points:
[tex]\[ f''\left(\frac{1 + 2\sqrt{7}}{3}\right) = 6\left(\frac{1 + 2\sqrt{7}}{3}\right) - 2 = 2 + 4\sqrt{7} > 0 \][/tex]
[tex]\[ f''\left(\frac{1 - 2\sqrt{7}}{3}\right) = 6\left(\frac{1 - 2\sqrt{7}}{3}\right) - 2 = 2 - 4\sqrt{7} < 0 \][/tex]
Since the second derivative changes sign at these critical points, [tex]\( f(x) \)[/tex] has both a local minimum and a local maximum. Therefore, the function is not one-to-one because it does not maintain a consistent increasing or decreasing trend across all values of [tex]\( x \)[/tex].
### Conclusion
Since [tex]\( f(x) \)[/tex] passes the vertical line test and has multiple y-values for some x-values (due to local maxima and minima), the function is many-to-one.
The correct answer is:
C. It is a many-to-one function.
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