Answer :
To answer the question of which exponential function grows at a faster rate than the quadratic function for [tex]\(0 < x < 3\)[/tex], we will analyze given data points and derive the quadratic function, then compare it with an exponential function.
### Step 1: Finding the Quadratic Function
The given data points are:
[tex]\( x = [0, 1, 2, 3] \)[/tex]
[tex]\( y = [0, 3, 12, 27] \)[/tex]
We will use these data points to form a quadratic equation of the form:
[tex]\[ y = ax^2 + bx + c \][/tex]
From the given data, we can deduce the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] for the quadratic function.
The coefficients are found to be:
[tex]\( a \approx 3.0 \)[/tex]
[tex]\( b \approx 0 \)[/tex]
[tex]\( c \approx 0 \)[/tex]
So, the quadratic function is:
[tex]\[ y = 3x^2 \][/tex]
### Step 2: Evaluation of Quadratic Function
We will evaluate the values of the quadratic function [tex]\( y = 3x^2 \)[/tex] for several points within the range [tex]\(0 \leq x \leq 3\)[/tex]:
For [tex]\( x \)[/tex] ranging from 0 to 3:
[tex]\[ \begin{align*} y(0) &= 3(0)^2 = 0 \\ y(0.5) &= 3(0.5)^2 = 0.75 \\ y(1) &= 3(1)^2 = 3 \\ y(1.5) &= 3(1.5)^2 = 6.75 \\ y(2) &= 3(2)^2 = 12 \\ y(2.5) &= 3(2.5)^2 = 18.75 \\ y(3) &= 3(3)^2 = 27 \\ \end{align*} \][/tex]
### Step 3: Evaluation of Exponential Function
Consider an exponential function of the form:
[tex]\[ y = e^x \][/tex]
Where [tex]\( e \approx 2.71828 \)[/tex].
Evaluate this function at several points within [tex]\(0 \leq x \leq 3\)[/tex]:
For [tex]\( x \)[/tex] ranging from 0 to 3:
[tex]\[ \begin{align*} y(0) &= e^{0} = 1 \\ y(0.5) &= e^{0.5} \approx 1.64872 \\ y(1) &= e^{1} \approx 2.71828 \\ y(1.5) &= e^{1.5} \approx 4.48169 \\ y(2) &= e^{2} \approx 7.38906 \\ y(2.5) &= e^{2.5} \approx 12.18249 \\ y(3) &= e^{3} \approx 20.08554 \\ \end{align*} \][/tex]
### Step 4: Comparison of Growth Rates
Now we compare the values of the quadratic function [tex]\( y = 3x^2 \)[/tex] and the exponential function [tex]\( y = e^x \)[/tex] over the range [tex]\( 0 \leq x \leq 3 \)[/tex].
[tex]\[ \begin{aligned} &\text{Quadratic Function:} & \quad & \text{Exponential Function:} \\ x & \quad y = 3x^2 & \quad & \quad y = e^x \\ 0 & \quad 0 & \quad & \quad 1 \\ 0.5 & \quad 0.75 & \quad & \quad 1.64872 \\ 1 & \quad 3 & \quad & \quad 2.71828 \\ 1.5 & \quad 6.75 & \quad & \quad 4.48169 \\ 2 & \quad 12 & \quad & \quad 7.38906 \\ 2.5 & \quad 18.75 & \quad & \quad 12.18249 \\ 3 & \quad 27 & \quad & \quad 20.08554 \\ \end{aligned} \][/tex]
From this comparison, we observe:
- At [tex]\( x = 0 \)[/tex], the exponential function starts at 1, while the quadratic function starts at 0.
- As [tex]\( x \)[/tex] increases from 0 to 3, the exponential function [tex]\( y = e^x \)[/tex] grows more rapidly than the quadratic function [tex]\( y = 3x^2 \)[/tex].
Therefore, the exponential function [tex]\( y = e^x \)[/tex] grows at a faster rate than the quadratic function [tex]\( y = 3x^2 \)[/tex] for [tex]\( 0 < x < 3 \)[/tex].
### Step 1: Finding the Quadratic Function
The given data points are:
[tex]\( x = [0, 1, 2, 3] \)[/tex]
[tex]\( y = [0, 3, 12, 27] \)[/tex]
We will use these data points to form a quadratic equation of the form:
[tex]\[ y = ax^2 + bx + c \][/tex]
From the given data, we can deduce the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] for the quadratic function.
The coefficients are found to be:
[tex]\( a \approx 3.0 \)[/tex]
[tex]\( b \approx 0 \)[/tex]
[tex]\( c \approx 0 \)[/tex]
So, the quadratic function is:
[tex]\[ y = 3x^2 \][/tex]
### Step 2: Evaluation of Quadratic Function
We will evaluate the values of the quadratic function [tex]\( y = 3x^2 \)[/tex] for several points within the range [tex]\(0 \leq x \leq 3\)[/tex]:
For [tex]\( x \)[/tex] ranging from 0 to 3:
[tex]\[ \begin{align*} y(0) &= 3(0)^2 = 0 \\ y(0.5) &= 3(0.5)^2 = 0.75 \\ y(1) &= 3(1)^2 = 3 \\ y(1.5) &= 3(1.5)^2 = 6.75 \\ y(2) &= 3(2)^2 = 12 \\ y(2.5) &= 3(2.5)^2 = 18.75 \\ y(3) &= 3(3)^2 = 27 \\ \end{align*} \][/tex]
### Step 3: Evaluation of Exponential Function
Consider an exponential function of the form:
[tex]\[ y = e^x \][/tex]
Where [tex]\( e \approx 2.71828 \)[/tex].
Evaluate this function at several points within [tex]\(0 \leq x \leq 3\)[/tex]:
For [tex]\( x \)[/tex] ranging from 0 to 3:
[tex]\[ \begin{align*} y(0) &= e^{0} = 1 \\ y(0.5) &= e^{0.5} \approx 1.64872 \\ y(1) &= e^{1} \approx 2.71828 \\ y(1.5) &= e^{1.5} \approx 4.48169 \\ y(2) &= e^{2} \approx 7.38906 \\ y(2.5) &= e^{2.5} \approx 12.18249 \\ y(3) &= e^{3} \approx 20.08554 \\ \end{align*} \][/tex]
### Step 4: Comparison of Growth Rates
Now we compare the values of the quadratic function [tex]\( y = 3x^2 \)[/tex] and the exponential function [tex]\( y = e^x \)[/tex] over the range [tex]\( 0 \leq x \leq 3 \)[/tex].
[tex]\[ \begin{aligned} &\text{Quadratic Function:} & \quad & \text{Exponential Function:} \\ x & \quad y = 3x^2 & \quad & \quad y = e^x \\ 0 & \quad 0 & \quad & \quad 1 \\ 0.5 & \quad 0.75 & \quad & \quad 1.64872 \\ 1 & \quad 3 & \quad & \quad 2.71828 \\ 1.5 & \quad 6.75 & \quad & \quad 4.48169 \\ 2 & \quad 12 & \quad & \quad 7.38906 \\ 2.5 & \quad 18.75 & \quad & \quad 12.18249 \\ 3 & \quad 27 & \quad & \quad 20.08554 \\ \end{aligned} \][/tex]
From this comparison, we observe:
- At [tex]\( x = 0 \)[/tex], the exponential function starts at 1, while the quadratic function starts at 0.
- As [tex]\( x \)[/tex] increases from 0 to 3, the exponential function [tex]\( y = e^x \)[/tex] grows more rapidly than the quadratic function [tex]\( y = 3x^2 \)[/tex].
Therefore, the exponential function [tex]\( y = e^x \)[/tex] grows at a faster rate than the quadratic function [tex]\( y = 3x^2 \)[/tex] for [tex]\( 0 < x < 3 \)[/tex].
