Find the best solutions to your problems with the help of IDNLearn.com's experts. Ask your questions and get detailed, reliable answers from our community of experienced experts.
Sagot :
To determine where the function [tex]\( f(x) \)[/tex] is increasing, we need to examine the given [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex] values.
Here is the table we have:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -6 & 34 \\ \hline -5 & 3 \\ \hline -4 & -10 \\ \hline -3 & -11 \\ \hline -2 & -6 \\ \hline -1 & -1 \\ \hline 0 & -2 \\ \hline 1 & -15 \\ \hline \end{array} \][/tex]
We look for intervals where [tex]\( f(x) \)[/tex] increases as [tex]\( x \)[/tex] increases.
1. From [tex]\( x = -6 \)[/tex] to [tex]\( x = -5 \)[/tex]; [tex]\( f(-6) = 34 \)[/tex] and [tex]\( f(-5) = 3 \)[/tex]
- [tex]\( f(-6) > f(-5) \)[/tex] (decreasing)
2. From [tex]\( x = -5 \)[/tex] to [tex]\( x = -4 \)[/tex]; [tex]\( f(-5) = 3 \)[/tex] and [tex]\( f(-4) = -10 \)[/tex]
- [tex]\( f(-5) > f(-4) \)[/tex] (decreasing)
3. From [tex]\( x = -4 \)[/tex] to [tex]\( x = -3 \)[/tex]; [tex]\( f(-4) = -10 \)[/tex] and [tex]\( f(-3) = -11 \)[/tex]
- [tex]\( f(-4) > f(-3) \)[/tex] (decreasing)
4. From [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]; [tex]\( f(-3) = -11 \)[/tex] and [tex]\( f(-2) = -6 \)[/tex]
- [tex]\( f(-3) < f(-2) \)[/tex] (increasing)
5. From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]; [tex]\( f(-2) = -6 \)[/tex] and [tex]\( f(-1) = -1 \)[/tex]
- [tex]\( f(-2) < f(-1) \)[/tex] (increasing)
6. From [tex]\( x = 0 \)[/tex]; [tex]\( f(0) = -2 \)[/tex]
- [tex]\( x = -1 \)[/tex] to [tex]\( x = 0; f(-1) = -1 \begins an increase and \(f(0)\ decreases decreasing 7. From \( x = 1) ; f(0) = -1; and \( f(1 = -15, 0) \ commences an increase and oint Checking each \( x \)[/tex]:
- The interval from \( x(-3, -2) \( -3 and \(-2( increasing becauseif (\( f(-3)\ <\ f(-2) )
- The interval \( x(-2, -1) \( from becauseif \( f(-2 incrementing -1)
Thus, the function \( f(x) is only between \( -3 -2 and \( -2 -1 respectively.
So, the correct interval where the function is increasing is:
\(\boxed{-3, -1}).
Here is the table we have:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -6 & 34 \\ \hline -5 & 3 \\ \hline -4 & -10 \\ \hline -3 & -11 \\ \hline -2 & -6 \\ \hline -1 & -1 \\ \hline 0 & -2 \\ \hline 1 & -15 \\ \hline \end{array} \][/tex]
We look for intervals where [tex]\( f(x) \)[/tex] increases as [tex]\( x \)[/tex] increases.
1. From [tex]\( x = -6 \)[/tex] to [tex]\( x = -5 \)[/tex]; [tex]\( f(-6) = 34 \)[/tex] and [tex]\( f(-5) = 3 \)[/tex]
- [tex]\( f(-6) > f(-5) \)[/tex] (decreasing)
2. From [tex]\( x = -5 \)[/tex] to [tex]\( x = -4 \)[/tex]; [tex]\( f(-5) = 3 \)[/tex] and [tex]\( f(-4) = -10 \)[/tex]
- [tex]\( f(-5) > f(-4) \)[/tex] (decreasing)
3. From [tex]\( x = -4 \)[/tex] to [tex]\( x = -3 \)[/tex]; [tex]\( f(-4) = -10 \)[/tex] and [tex]\( f(-3) = -11 \)[/tex]
- [tex]\( f(-4) > f(-3) \)[/tex] (decreasing)
4. From [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]; [tex]\( f(-3) = -11 \)[/tex] and [tex]\( f(-2) = -6 \)[/tex]
- [tex]\( f(-3) < f(-2) \)[/tex] (increasing)
5. From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]; [tex]\( f(-2) = -6 \)[/tex] and [tex]\( f(-1) = -1 \)[/tex]
- [tex]\( f(-2) < f(-1) \)[/tex] (increasing)
6. From [tex]\( x = 0 \)[/tex]; [tex]\( f(0) = -2 \)[/tex]
- [tex]\( x = -1 \)[/tex] to [tex]\( x = 0; f(-1) = -1 \begins an increase and \(f(0)\ decreases decreasing 7. From \( x = 1) ; f(0) = -1; and \( f(1 = -15, 0) \ commences an increase and oint Checking each \( x \)[/tex]:
- The interval from \( x(-3, -2) \( -3 and \(-2( increasing becauseif (\( f(-3)\ <\ f(-2) )
- The interval \( x(-2, -1) \( from becauseif \( f(-2 incrementing -1)
Thus, the function \( f(x) is only between \( -3 -2 and \( -2 -1 respectively.
So, the correct interval where the function is increasing is:
\(\boxed{-3, -1}).
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. Find clear and concise answers at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.