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To determine which statement about the function [tex]\( f(x) = -(x+6)(x+2) \)[/tex] is true, let's analyze the function step by step:
1. Expand the function:
[tex]\( f(x) = -(x+6)(x+2) \)[/tex]
Expanding the product inside the parentheses:
[tex]\( f(x) = -(x^2 + 8x + 12) \)[/tex]
Simplify by distributing the negative sign:
[tex]\( f(x) = -x^2 - 8x - 12 \)[/tex]
2. Find the derivative of the function to determine its critical points and behavior:
The derivative [tex]\( f'(x) \)[/tex] is calculated as follows:
[tex]\( f'(x) = \frac{d}{dx} (-x^2 - 8x - 12) \)[/tex]
Using the power rule:
[tex]\( f'(x) = -2x - 8 \)[/tex]
3. Set the derivative equal to zero to find critical points:
[tex]\( -2x - 8 = 0 \)[/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\( -2x = 8 \)[/tex]
[tex]\( x = -4 \)[/tex]
The critical point is [tex]\( x = -4 \)[/tex].
4. Determine whether the function is increasing or decreasing around the critical point:
- For [tex]\( x < -4 \)[/tex], let's choose [tex]\( x = -5 \)[/tex]:
[tex]\[ f'(-5) = -2(-5) - 8 = 10 - 8 = 2 \][/tex]
Since [tex]\( f'(-5) > 0 \)[/tex], the function is increasing for [tex]\( x < -4 \)[/tex].
- For [tex]\( x > -4 \)[/tex], let's choose [tex]\( x = -3 \)[/tex]:
[tex]\[ f'(-3) = -2(-3) - 8 = 6 - 8 = -2 \][/tex]
Since [tex]\( f'(-3) < 0 \)[/tex], the function is decreasing for [tex]\( x > -4 \)[/tex].
5. Evaluate the given statements:
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
This is true because we determined that the derivative [tex]\( f'(x) \)[/tex] is positive for [tex]\( x < -4 \)[/tex].
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( -6 < x < -2 \)[/tex].
This is false because the function is not increasing in the interval [tex]\( -6 < x < -2 \)[/tex], especially since it decreases for [tex]\( x > -4 \)[/tex].
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -6 \)[/tex] and where [tex]\( x > -2 \)[/tex].
This is false because we have established that the function is increasing for [tex]\( x < -4 \)[/tex].
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
This is false as the function is increasing in that interval.
Therefore, the correct statement is:
The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
Thus, the correct answer is the first statement.
1. Expand the function:
[tex]\( f(x) = -(x+6)(x+2) \)[/tex]
Expanding the product inside the parentheses:
[tex]\( f(x) = -(x^2 + 8x + 12) \)[/tex]
Simplify by distributing the negative sign:
[tex]\( f(x) = -x^2 - 8x - 12 \)[/tex]
2. Find the derivative of the function to determine its critical points and behavior:
The derivative [tex]\( f'(x) \)[/tex] is calculated as follows:
[tex]\( f'(x) = \frac{d}{dx} (-x^2 - 8x - 12) \)[/tex]
Using the power rule:
[tex]\( f'(x) = -2x - 8 \)[/tex]
3. Set the derivative equal to zero to find critical points:
[tex]\( -2x - 8 = 0 \)[/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\( -2x = 8 \)[/tex]
[tex]\( x = -4 \)[/tex]
The critical point is [tex]\( x = -4 \)[/tex].
4. Determine whether the function is increasing or decreasing around the critical point:
- For [tex]\( x < -4 \)[/tex], let's choose [tex]\( x = -5 \)[/tex]:
[tex]\[ f'(-5) = -2(-5) - 8 = 10 - 8 = 2 \][/tex]
Since [tex]\( f'(-5) > 0 \)[/tex], the function is increasing for [tex]\( x < -4 \)[/tex].
- For [tex]\( x > -4 \)[/tex], let's choose [tex]\( x = -3 \)[/tex]:
[tex]\[ f'(-3) = -2(-3) - 8 = 6 - 8 = -2 \][/tex]
Since [tex]\( f'(-3) < 0 \)[/tex], the function is decreasing for [tex]\( x > -4 \)[/tex].
5. Evaluate the given statements:
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
This is true because we determined that the derivative [tex]\( f'(x) \)[/tex] is positive for [tex]\( x < -4 \)[/tex].
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( -6 < x < -2 \)[/tex].
This is false because the function is not increasing in the interval [tex]\( -6 < x < -2 \)[/tex], especially since it decreases for [tex]\( x > -4 \)[/tex].
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -6 \)[/tex] and where [tex]\( x > -2 \)[/tex].
This is false because we have established that the function is increasing for [tex]\( x < -4 \)[/tex].
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
This is false as the function is increasing in that interval.
Therefore, the correct statement is:
The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -4 \)[/tex].
Thus, the correct answer is the first statement.
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