Answer :
To analyze which exponential function grows at a faster rate than the given quadratic function within the interval [tex]\( 0 < x < 3 \)[/tex], we need to compare the rates of change of both functions over the given range.
First, let's calculate the rate of change of the quadratic function:
Given values for the quadratic function:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 0 \\ \hline 1 & 3 \\ \hline 2 & 12 \\ \hline 3 & 27 \\ \hline \end{array} \][/tex]
The rate of change over each interval [tex]\([x_i, x_{i+1}]\)[/tex] can be calculated using the formula:
[tex]\[ \text{Rate of change} = \frac{y_{i+1} - y_i}{x_{i+1} - x_i} \][/tex]
For the interval [tex]\([0, 1]\)[/tex]:
[tex]\[ \text{Rate of change}_1 = \frac{3 - 0}{1 - 0} = 3.0 \][/tex]
For the interval [tex]\([1, 2]\)[/tex]:
[tex]\[ \text{Rate of change}_2 = \frac{12 - 3}{2 - 1} = 9.0 \][/tex]
For the interval [tex]\([2, 3]\)[/tex]:
[tex]\[ \text{Rate of change}_3 = \frac{27 - 12}{3 - 2} = 15.0 \][/tex]
So, the rates of change for the quadratic function are:
[tex]\[ [3.0, 9.0, 15.0] \][/tex]
Next, consider an exponential function of the general form [tex]\( y = a \cdot b^x \)[/tex]. For simplicity, let's analyze the exponential function [tex]\( y = 2^x \)[/tex].
Given values for the exponential function [tex]\( y = 2^x \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 1 & 2 \\ \hline 2 & 4 \\ \hline 3 & 8 \\ \hline \end{array} \][/tex]
Again, we calculate the rate of change over each interval:
For the interval [tex]\([0, 1]\)[/tex]:
[tex]\[ \text{Rate of change}_1 = \frac{2 - 1}{1 - 0} = 1.0 \][/tex]
For the interval [tex]\([1, 2]\)[/tex]:
[tex]\[ \text{Rate of change}_2 = \frac{4 - 2}{2 - 1} = 2.0 \][/tex]
For the interval [tex]\([2, 3]\)[/tex]:
[tex]\[ \text{Rate of change}_3 = \frac{8 - 4}{3 - 2} = 4.0 \][/tex]
So, the rates of change for the exponential function [tex]\( y = 2^x \)[/tex] are:
[tex]\[ [1.0, 2.0, 4.0] \][/tex]
Comparing the rates of change:
- Quadratic function: [tex]\([3.0, 9.0, 15.0]\)[/tex]
- Exponential function [tex]\( y = 2^x \)[/tex]: [tex]\([1.0, 2.0, 4.0]\)[/tex]
Clearly, the quadratic function has higher rates of change within the interval [tex]\( 0 < x < 3 \)[/tex] compared to the exponential function [tex]\( y = 2^x \)[/tex].
To find an exponential function that grows faster, we need an exponential function where the rate of change exceeds 15.0 by [tex]\( x = 3 \)[/tex]. Let's consider a stronger exponential function [tex]\( y = 4^x \)[/tex]:
Given values for the exponential function [tex]\( y = 4^x \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 1 & 4 \\ \hline 2 & 16 \\ \hline 3 & 64 \\ \hline \end{array} \][/tex]
Now, calculate the rate of change:
For the interval [tex]\([0, 1]\)[/tex]:
[tex]\[ \text{Rate of change}_1 = \frac{4 - 1}{1 - 0} = 3.0 \][/tex]
For the interval [tex]\([1, 2]\)[/tex]:
[tex]\[ \text{Rate of change}_2 = \frac{16 - 4}{2 - 1} = 12.0 \][/tex]
For the interval [tex]\([2, 3]\)[/tex]:
[tex]\[ \text{Rate of change}_3 = \frac{64 - 16}{3 - 2} = 48.0 \][/tex]
So, the rates of change for the exponential function [tex]\( y = 4^x \)[/tex] are:
[tex]\[ [3.0, 12.0, 48.0] \][/tex]
When we compare:
- Quadratic function: [tex]\([3.0, 9.0, 15.0]\)[/tex]
- Exponential function [tex]\( y = 4^x \)[/tex]: [tex]\([3.0, 12.0, 48.0]\)[/tex]
It is evident that the exponential function [tex]\( y = 4^x \)[/tex] eventually grows faster than the quadratic function. Since its rate of change surpasses that of the quadratic function in the interval [tex]\( 0 < x < 3 \)[/tex], we conclude that the exponential function [tex]\( y = 4^x \)[/tex] grows at a faster rate than the given quadratic function in the specified interval.
First, let's calculate the rate of change of the quadratic function:
Given values for the quadratic function:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 0 \\ \hline 1 & 3 \\ \hline 2 & 12 \\ \hline 3 & 27 \\ \hline \end{array} \][/tex]
The rate of change over each interval [tex]\([x_i, x_{i+1}]\)[/tex] can be calculated using the formula:
[tex]\[ \text{Rate of change} = \frac{y_{i+1} - y_i}{x_{i+1} - x_i} \][/tex]
For the interval [tex]\([0, 1]\)[/tex]:
[tex]\[ \text{Rate of change}_1 = \frac{3 - 0}{1 - 0} = 3.0 \][/tex]
For the interval [tex]\([1, 2]\)[/tex]:
[tex]\[ \text{Rate of change}_2 = \frac{12 - 3}{2 - 1} = 9.0 \][/tex]
For the interval [tex]\([2, 3]\)[/tex]:
[tex]\[ \text{Rate of change}_3 = \frac{27 - 12}{3 - 2} = 15.0 \][/tex]
So, the rates of change for the quadratic function are:
[tex]\[ [3.0, 9.0, 15.0] \][/tex]
Next, consider an exponential function of the general form [tex]\( y = a \cdot b^x \)[/tex]. For simplicity, let's analyze the exponential function [tex]\( y = 2^x \)[/tex].
Given values for the exponential function [tex]\( y = 2^x \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 1 & 2 \\ \hline 2 & 4 \\ \hline 3 & 8 \\ \hline \end{array} \][/tex]
Again, we calculate the rate of change over each interval:
For the interval [tex]\([0, 1]\)[/tex]:
[tex]\[ \text{Rate of change}_1 = \frac{2 - 1}{1 - 0} = 1.0 \][/tex]
For the interval [tex]\([1, 2]\)[/tex]:
[tex]\[ \text{Rate of change}_2 = \frac{4 - 2}{2 - 1} = 2.0 \][/tex]
For the interval [tex]\([2, 3]\)[/tex]:
[tex]\[ \text{Rate of change}_3 = \frac{8 - 4}{3 - 2} = 4.0 \][/tex]
So, the rates of change for the exponential function [tex]\( y = 2^x \)[/tex] are:
[tex]\[ [1.0, 2.0, 4.0] \][/tex]
Comparing the rates of change:
- Quadratic function: [tex]\([3.0, 9.0, 15.0]\)[/tex]
- Exponential function [tex]\( y = 2^x \)[/tex]: [tex]\([1.0, 2.0, 4.0]\)[/tex]
Clearly, the quadratic function has higher rates of change within the interval [tex]\( 0 < x < 3 \)[/tex] compared to the exponential function [tex]\( y = 2^x \)[/tex].
To find an exponential function that grows faster, we need an exponential function where the rate of change exceeds 15.0 by [tex]\( x = 3 \)[/tex]. Let's consider a stronger exponential function [tex]\( y = 4^x \)[/tex]:
Given values for the exponential function [tex]\( y = 4^x \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 1 \\ \hline 1 & 4 \\ \hline 2 & 16 \\ \hline 3 & 64 \\ \hline \end{array} \][/tex]
Now, calculate the rate of change:
For the interval [tex]\([0, 1]\)[/tex]:
[tex]\[ \text{Rate of change}_1 = \frac{4 - 1}{1 - 0} = 3.0 \][/tex]
For the interval [tex]\([1, 2]\)[/tex]:
[tex]\[ \text{Rate of change}_2 = \frac{16 - 4}{2 - 1} = 12.0 \][/tex]
For the interval [tex]\([2, 3]\)[/tex]:
[tex]\[ \text{Rate of change}_3 = \frac{64 - 16}{3 - 2} = 48.0 \][/tex]
So, the rates of change for the exponential function [tex]\( y = 4^x \)[/tex] are:
[tex]\[ [3.0, 12.0, 48.0] \][/tex]
When we compare:
- Quadratic function: [tex]\([3.0, 9.0, 15.0]\)[/tex]
- Exponential function [tex]\( y = 4^x \)[/tex]: [tex]\([3.0, 12.0, 48.0]\)[/tex]
It is evident that the exponential function [tex]\( y = 4^x \)[/tex] eventually grows faster than the quadratic function. Since its rate of change surpasses that of the quadratic function in the interval [tex]\( 0 < x < 3 \)[/tex], we conclude that the exponential function [tex]\( y = 4^x \)[/tex] grows at a faster rate than the given quadratic function in the specified interval.
