Discover how IDNLearn.com can help you find the answers you need quickly and easily. Our platform is designed to provide trustworthy and thorough answers to any questions you may have.
Sagot :
To find the largest interval of [tex]\( x \)[/tex] values where the function [tex]\( f(x) = -x^3 + 4x + 3 \)[/tex] is increasing, we need to examine the given values of [tex]\( f(x) \)[/tex] provided in the table:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -3 & 18 \\ \hline -2 & 3 \\ \hline -1 & 0 \\ \hline 0 & 3 \\ \hline 1 & 6 \\ \hline 2 & 3 \\ \hline \end{array} \][/tex]
Next, we calculate the differences between successive [tex]\( f(x) \)[/tex] values to determine where the function is increasing.
1. Calculate differences between successive [tex]\( f(x) \)[/tex] values:
[tex]\[ \Delta f(x) = f(x_{i+1}) - f(x_i) \][/tex]
- For [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) - f(-3) = 3 - 18 = -15 \][/tex]
- For [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) - f(-2) = 0 - 3 = -3 \][/tex]
- For [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) - f(-1) = 3 - 0 = 3 \][/tex]
- For [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) - f(0) = 6 - 3 = 3 \][/tex]
- For [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) - f(1) = 3 - 6 = -3 \][/tex]
2. Identify the intervals where the differences are positive, indicating the function is increasing:
[tex]\[ \begin{aligned} & \text{Interval from } x = -3 \text{ to } x = -2: \Delta f(x) = -15 \quad (\text{decreasing}) \\ & \text{Interval from } x = -2 \text{ to } x = -1: \Delta f(x) = -3 \quad (\text{decreasing}) \\ & \text{Interval from } x = -1 \text{ to } x = 0: \Delta f(x) = 3 \quad (\text{increasing}) \\ & \text{Interval from } x = 0 \text{ to } x = 1: \Delta f(x) = 3 \quad (\text{increasing}) \\ & \text{Interval from } x = 1 \text{ to } x = 2: \Delta f(x) = -3 \quad (\text{decreasing}) \\ \end{aligned} \][/tex]
From these differences, the intervals where the function is increasing are:
[tex]\[ (-1, 0) \text{ and } (0, 1) \][/tex]
Thus, combining these, we conclude the largest interval where the function is increasing is:
[tex]\[ (-1, 1) \][/tex]
So, the largest interval of [tex]\( x \)[/tex] values where the function [tex]\( f(x) \)[/tex] is increasing is:
[tex]\[ \boxed{(-1, 1)} \][/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -3 & 18 \\ \hline -2 & 3 \\ \hline -1 & 0 \\ \hline 0 & 3 \\ \hline 1 & 6 \\ \hline 2 & 3 \\ \hline \end{array} \][/tex]
Next, we calculate the differences between successive [tex]\( f(x) \)[/tex] values to determine where the function is increasing.
1. Calculate differences between successive [tex]\( f(x) \)[/tex] values:
[tex]\[ \Delta f(x) = f(x_{i+1}) - f(x_i) \][/tex]
- For [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) - f(-3) = 3 - 18 = -15 \][/tex]
- For [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) - f(-2) = 0 - 3 = -3 \][/tex]
- For [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) - f(-1) = 3 - 0 = 3 \][/tex]
- For [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) - f(0) = 6 - 3 = 3 \][/tex]
- For [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) - f(1) = 3 - 6 = -3 \][/tex]
2. Identify the intervals where the differences are positive, indicating the function is increasing:
[tex]\[ \begin{aligned} & \text{Interval from } x = -3 \text{ to } x = -2: \Delta f(x) = -15 \quad (\text{decreasing}) \\ & \text{Interval from } x = -2 \text{ to } x = -1: \Delta f(x) = -3 \quad (\text{decreasing}) \\ & \text{Interval from } x = -1 \text{ to } x = 0: \Delta f(x) = 3 \quad (\text{increasing}) \\ & \text{Interval from } x = 0 \text{ to } x = 1: \Delta f(x) = 3 \quad (\text{increasing}) \\ & \text{Interval from } x = 1 \text{ to } x = 2: \Delta f(x) = -3 \quad (\text{decreasing}) \\ \end{aligned} \][/tex]
From these differences, the intervals where the function is increasing are:
[tex]\[ (-1, 0) \text{ and } (0, 1) \][/tex]
Thus, combining these, we conclude the largest interval where the function is increasing is:
[tex]\[ (-1, 1) \][/tex]
So, the largest interval of [tex]\( x \)[/tex] values where the function [tex]\( f(x) \)[/tex] is increasing is:
[tex]\[ \boxed{(-1, 1)} \][/tex]
Your participation is crucial to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.
