Answer :
To determine which statement about the function [tex]\( f(x) = (x-4)(x+1) \)[/tex] is true, we will analyze the behavior of the function, particularly whether it is increasing or decreasing over different intervals. We start by finding the derivative of the function to determine the intervals of increase and decrease.
1. Finding the derivative of [tex]\( f(x) \)[/tex]:
We begin by expanding the function:
[tex]\[ f(x) = (x-4)(x+1) = x^2 + x - 4x - 4 = x^2 - 3x - 4. \][/tex]
Now, take the derivative:
[tex]\[ f'(x) = \frac{d}{dx}(x^2 - 3x - 4) = 2x - 3. \][/tex]
2. Finding critical points:
Critical points occur where the derivative is zero or undefined. Since the derivative [tex]\( f'(x) = 2x - 3 \)[/tex] is a linear function and defined everywhere, we find the critical points by setting the derivative to zero:
[tex]\[ 2x - 3 = 0 \implies x = \frac{3}{2}. \][/tex]
3. Determining the sign of [tex]\( f'(x) \)[/tex] around the critical point:
We need to check the sign of [tex]\( f'(x) \)[/tex] to determine whether [tex]\( f(x) \)[/tex] is increasing or decreasing on the intervals formed by the critical point [tex]\( x = \frac{3}{2} \)[/tex].
- For [tex]\( x < \frac{3}{2} \)[/tex]:
[tex]\[ \text{Choose a test point, e.g., } x = 0: \quad f'(0) = 2(0) - 3 = -3. \][/tex]
Since [tex]\( f'(0) < 0 \)[/tex], the function [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x < \frac{3}{2} \)[/tex].
- For [tex]\( x > \frac{3}{2} \)[/tex]:
[tex]\[ \text{Choose a test point, e.g., } x = 2: \quad f'(2) = 2(2) - 3 = 1. \][/tex]
Since [tex]\( f'(2) > 0 \)[/tex], the function [tex]\( f(x) \)[/tex] is increasing for [tex]\( x > \frac{3}{2} \)[/tex].
4. Combining the results:
- The function [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x < \frac{3}{2} \)[/tex].
- The function [tex]\( f(x) \)[/tex] is increasing for [tex]\( x > \frac{3}{2} \)[/tex].
5. Verifying statements:
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < 0 \)[/tex]:
This statement is false since [tex]\( x < 0 \)[/tex] is within the decreasing interval.
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -1 \)[/tex] and where [tex]\( x > 4 \)[/tex]:
This statement is false. The correct increasing interval is [tex]\( x > \frac{3}{2} \)[/tex].
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( -1 < x < 4 \)[/tex]:
This statement is false since [tex]\( \frac{3}{2} \)[/tex] is within the interval [tex]\([ -1, 4 ]\)[/tex] and marks the transition from decreasing to increasing behavior.
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < 1.5 \)[/tex]:
This statement is true since [tex]\( 1.5 \)[/tex] is equivalent to [tex]\( \frac{3}{2} \)[/tex], marking the point where the function changes from decreasing to increasing.
Thus, the correct statement is:
[tex]\[ \text{The function is decreasing for all real values of } x \text{ where } x < 1.5. \][/tex]
1. Finding the derivative of [tex]\( f(x) \)[/tex]:
We begin by expanding the function:
[tex]\[ f(x) = (x-4)(x+1) = x^2 + x - 4x - 4 = x^2 - 3x - 4. \][/tex]
Now, take the derivative:
[tex]\[ f'(x) = \frac{d}{dx}(x^2 - 3x - 4) = 2x - 3. \][/tex]
2. Finding critical points:
Critical points occur where the derivative is zero or undefined. Since the derivative [tex]\( f'(x) = 2x - 3 \)[/tex] is a linear function and defined everywhere, we find the critical points by setting the derivative to zero:
[tex]\[ 2x - 3 = 0 \implies x = \frac{3}{2}. \][/tex]
3. Determining the sign of [tex]\( f'(x) \)[/tex] around the critical point:
We need to check the sign of [tex]\( f'(x) \)[/tex] to determine whether [tex]\( f(x) \)[/tex] is increasing or decreasing on the intervals formed by the critical point [tex]\( x = \frac{3}{2} \)[/tex].
- For [tex]\( x < \frac{3}{2} \)[/tex]:
[tex]\[ \text{Choose a test point, e.g., } x = 0: \quad f'(0) = 2(0) - 3 = -3. \][/tex]
Since [tex]\( f'(0) < 0 \)[/tex], the function [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x < \frac{3}{2} \)[/tex].
- For [tex]\( x > \frac{3}{2} \)[/tex]:
[tex]\[ \text{Choose a test point, e.g., } x = 2: \quad f'(2) = 2(2) - 3 = 1. \][/tex]
Since [tex]\( f'(2) > 0 \)[/tex], the function [tex]\( f(x) \)[/tex] is increasing for [tex]\( x > \frac{3}{2} \)[/tex].
4. Combining the results:
- The function [tex]\( f(x) \)[/tex] is decreasing for [tex]\( x < \frac{3}{2} \)[/tex].
- The function [tex]\( f(x) \)[/tex] is increasing for [tex]\( x > \frac{3}{2} \)[/tex].
5. Verifying statements:
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < 0 \)[/tex]:
This statement is false since [tex]\( x < 0 \)[/tex] is within the decreasing interval.
- The function is increasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < -1 \)[/tex] and where [tex]\( x > 4 \)[/tex]:
This statement is false. The correct increasing interval is [tex]\( x > \frac{3}{2} \)[/tex].
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( -1 < x < 4 \)[/tex]:
This statement is false since [tex]\( \frac{3}{2} \)[/tex] is within the interval [tex]\([ -1, 4 ]\)[/tex] and marks the transition from decreasing to increasing behavior.
- The function is decreasing for all real values of [tex]\( x \)[/tex] where [tex]\( x < 1.5 \)[/tex]:
This statement is true since [tex]\( 1.5 \)[/tex] is equivalent to [tex]\( \frac{3}{2} \)[/tex], marking the point where the function changes from decreasing to increasing.
Thus, the correct statement is:
[tex]\[ \text{The function is decreasing for all real values of } x \text{ where } x < 1.5. \][/tex]
