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To solve the problem regarding the function [tex]\( f(x) = (x-3)(x+1) \)[/tex] and describe the values for which the graph is both positive and decreasing, we need to analyze the behavior of the function in detail.
1. Identify the roots of the function [tex]\( f(x) = (x-3)(x+1) \)[/tex]:
- Set [tex]\( f(x) = 0 \)[/tex].
- [tex]\( (x-3)(x+1) = 0 \)[/tex].
- The roots are [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex].
2. Determine intervals where [tex]\( f(x) \)[/tex] is positive or negative:
- Interval [tex]\( x < -1 \)[/tex]:
- Both factors [tex]\( (x-3) \)[/tex] and [tex]\( (x+1) \)[/tex] are negative, so [tex]\( f(x) > 0 \)[/tex].
- Interval [tex]\( -1 < x < 3 \)[/tex]:
- Factor [tex]\( (x-3) \)[/tex] is negative and [tex]\( (x+1) \)[/tex] is positive in this interval, so [tex]\( f(x) < 0 \)[/tex].
- Interval [tex]\( x > 3 \)[/tex]:
- Both factors [tex]\( (x-3) \)[/tex] and [tex]\( (x+1) \)[/tex] are positive, so [tex]\( f(x) > 0 \)[/tex].
3. Calculate the derivative [tex]\( f'(x) \)[/tex] to find where the function is decreasing:
- Use the product rule: [tex]\( f(x) = (x-3)(x+1) \)[/tex].
- [tex]\( f'(x) = (x-3) \cdot 1 + (x+1) \cdot 1 = x-3 + x+1 = 2x - 2 \)[/tex].
4. Set the derivative [tex]\( f'(x) = 2x - 2 \)[/tex] to 0 to find critical points:
- [tex]\( 2x - 2 = 0 \)[/tex].
- [tex]\( 2x = 2 \)[/tex].
- [tex]\( x = 1 \)[/tex].
5. Determine where [tex]\( f'(x) \)[/tex] is negative (i.e., where the function is decreasing):
- For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) = 2x - 2 \)[/tex] is negative, implying [tex]\( f(x) \)[/tex] is decreasing.
- For [tex]\( x > 1 \)[/tex], [tex]\( f'(x) = 2x - 2 \)[/tex] is positive, implying [tex]\( f(x) \)[/tex] is increasing.
6. Combine the intervals for positivity and decreasing behavior:
- [tex]\( f(x) \)[/tex] is positive in intervals [tex]\( x < -1 \)[/tex] and [tex]\( x > 3 \)[/tex], but we are interested in where the function is also decreasing.
- [tex]\( f(x) \)[/tex] is decreasing when [tex]\( x < 1 \)[/tex].
- The overlap between the intervals where [tex]\( f(x) \)[/tex] is positive and decreasing is [tex]\( 1 < x < 3 \)[/tex].
Therefore, the values for which the graph of the function [tex]\( f(x) = (x-3)(x+1) \)[/tex] is positive and decreasing is:
[tex]\[ \boxed{1 < x < 3} \][/tex]
1. Identify the roots of the function [tex]\( f(x) = (x-3)(x+1) \)[/tex]:
- Set [tex]\( f(x) = 0 \)[/tex].
- [tex]\( (x-3)(x+1) = 0 \)[/tex].
- The roots are [tex]\( x = 3 \)[/tex] and [tex]\( x = -1 \)[/tex].
2. Determine intervals where [tex]\( f(x) \)[/tex] is positive or negative:
- Interval [tex]\( x < -1 \)[/tex]:
- Both factors [tex]\( (x-3) \)[/tex] and [tex]\( (x+1) \)[/tex] are negative, so [tex]\( f(x) > 0 \)[/tex].
- Interval [tex]\( -1 < x < 3 \)[/tex]:
- Factor [tex]\( (x-3) \)[/tex] is negative and [tex]\( (x+1) \)[/tex] is positive in this interval, so [tex]\( f(x) < 0 \)[/tex].
- Interval [tex]\( x > 3 \)[/tex]:
- Both factors [tex]\( (x-3) \)[/tex] and [tex]\( (x+1) \)[/tex] are positive, so [tex]\( f(x) > 0 \)[/tex].
3. Calculate the derivative [tex]\( f'(x) \)[/tex] to find where the function is decreasing:
- Use the product rule: [tex]\( f(x) = (x-3)(x+1) \)[/tex].
- [tex]\( f'(x) = (x-3) \cdot 1 + (x+1) \cdot 1 = x-3 + x+1 = 2x - 2 \)[/tex].
4. Set the derivative [tex]\( f'(x) = 2x - 2 \)[/tex] to 0 to find critical points:
- [tex]\( 2x - 2 = 0 \)[/tex].
- [tex]\( 2x = 2 \)[/tex].
- [tex]\( x = 1 \)[/tex].
5. Determine where [tex]\( f'(x) \)[/tex] is negative (i.e., where the function is decreasing):
- For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) = 2x - 2 \)[/tex] is negative, implying [tex]\( f(x) \)[/tex] is decreasing.
- For [tex]\( x > 1 \)[/tex], [tex]\( f'(x) = 2x - 2 \)[/tex] is positive, implying [tex]\( f(x) \)[/tex] is increasing.
6. Combine the intervals for positivity and decreasing behavior:
- [tex]\( f(x) \)[/tex] is positive in intervals [tex]\( x < -1 \)[/tex] and [tex]\( x > 3 \)[/tex], but we are interested in where the function is also decreasing.
- [tex]\( f(x) \)[/tex] is decreasing when [tex]\( x < 1 \)[/tex].
- The overlap between the intervals where [tex]\( f(x) \)[/tex] is positive and decreasing is [tex]\( 1 < x < 3 \)[/tex].
Therefore, the values for which the graph of the function [tex]\( f(x) = (x-3)(x+1) \)[/tex] is positive and decreasing is:
[tex]\[ \boxed{1 < x < 3} \][/tex]
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