Answer :
To address the given question, let's carefully examine the growth of three functions: [tex]\( y_1 = 2x \)[/tex], [tex]\( y_2 = 2x^2 \)[/tex], and [tex]\( y_3 = 2^x \)[/tex]. Our aim is to determine the [tex]\( y \)[/tex]-value at which the exponential function [tex]\( y_3 \)[/tex] surpasses the linear function [tex]\( y_1 \)[/tex] and the quadratic function [tex]\( y_2 \)[/tex], and to identify which of these functions grows the fastest.
### Comparison with the Linear Function [tex]\( y_1 = 2x \)[/tex]
First, we compare the exponential function [tex]\( y_3 = 2^x \)[/tex] with the linear function [tex]\( y_1 = 2x \)[/tex].
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ y_1 = 2 \cdot 0 = 0 \quad \text{and} \quad y_3 = 2^0 = 1 \][/tex]
Here, [tex]\( y_3 = 1 \)[/tex] when [tex]\( y_1 = 0 \)[/tex].
- Increase [tex]\( x \)[/tex] incrementally, we reach [tex]\( x = 1 \)[/tex]:
[tex]\[ y_1 = 2 \cdot 1 = 2 \quad \text{and} \quad y_3 = 2^1 = 2 \][/tex]
Here, [tex]\( y_3 = 2 \)[/tex], matching [tex]\( y_1 \)[/tex].
Continuing this process, the exponential function starts growing faster. The exponential function [tex]\( y_3 \)[/tex] surpasses the linear function [tex]\( y_1 \)[/tex] immediately after this point. Therefore, the [tex]\( y \)[/tex]-value at which [tex]\( y_3 \)[/tex] surpasses [tex]\( y_1 \)[/tex] as we increase [tex]\( x \)[/tex] is:
[tex]\[ \boxed{1} \][/tex]
### Comparison with the Quadratic Function [tex]\( y_2 = 2x^2 \)[/tex]
Next, we compare the exponential function [tex]\( y_3 = 2^x \)[/tex] with the quadratic function [tex]\( y_2 = 2x^2 \)[/tex].
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ y_2 = 2 \cdot 0^2 = 0 \quad \text{and} \quad y_3 = 2^0 = 1 \][/tex]
Here, [tex]\( y_3 = 1 \)[/tex] when [tex]\( y_2 = 0 \)[/tex].
- Increase [tex]\( x \)[/tex] incrementally, we reach [tex]\( x = 1 \)[/tex]:
[tex]\[ y_2 = 2 \cdot 1^2 = 2 \quad \text{and} \quad y_3 = 2^1 = 2 \][/tex]
Here, [tex]\( y_3 = 2 \)[/tex], matching [tex]\( y_2 \)[/tex].
As the values of [tex]\( x \)[/tex] increase further, the exponential function [tex]\( y_3 \)[/tex] starts to surpass the quadratic function [tex]\( y_2 \)[/tex]. Therefore, the [tex]\( y \)[/tex]-value at which [tex]\( y_3 \)[/tex] surpasses [tex]\( y_2 \)[/tex] immediately after this point is:
[tex]\[ \boxed{1} \][/tex]
### Identifying the Fastest Growing Function
Among the three functions [tex]\( y_1 = 2x \)[/tex], [tex]\( y_2 = 2x^2 \)[/tex], and [tex]\( y_3 = 2^x \)[/tex], the exponential function grows the fastest. This is because exponential growth increases more rapidly than both linear and quadratic growth as [tex]\( x \)[/tex] increases.
Thus, the function that grows the fastest is:
[tex]\[ \boxed{\text{the exponential function}} \][/tex]
In summary:
- The exponential function [tex]\( y_3 \)[/tex] surpasses the linear function [tex]\( y_1 \)[/tex] at [tex]\( y = 1 \)[/tex].
- The exponential function [tex]\( y_3 \)[/tex] surpasses the quadratic function [tex]\( y_2 \)[/tex] at [tex]\( y = 1 \)[/tex].
- The exponential function grows the fastest.
### Comparison with the Linear Function [tex]\( y_1 = 2x \)[/tex]
First, we compare the exponential function [tex]\( y_3 = 2^x \)[/tex] with the linear function [tex]\( y_1 = 2x \)[/tex].
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ y_1 = 2 \cdot 0 = 0 \quad \text{and} \quad y_3 = 2^0 = 1 \][/tex]
Here, [tex]\( y_3 = 1 \)[/tex] when [tex]\( y_1 = 0 \)[/tex].
- Increase [tex]\( x \)[/tex] incrementally, we reach [tex]\( x = 1 \)[/tex]:
[tex]\[ y_1 = 2 \cdot 1 = 2 \quad \text{and} \quad y_3 = 2^1 = 2 \][/tex]
Here, [tex]\( y_3 = 2 \)[/tex], matching [tex]\( y_1 \)[/tex].
Continuing this process, the exponential function starts growing faster. The exponential function [tex]\( y_3 \)[/tex] surpasses the linear function [tex]\( y_1 \)[/tex] immediately after this point. Therefore, the [tex]\( y \)[/tex]-value at which [tex]\( y_3 \)[/tex] surpasses [tex]\( y_1 \)[/tex] as we increase [tex]\( x \)[/tex] is:
[tex]\[ \boxed{1} \][/tex]
### Comparison with the Quadratic Function [tex]\( y_2 = 2x^2 \)[/tex]
Next, we compare the exponential function [tex]\( y_3 = 2^x \)[/tex] with the quadratic function [tex]\( y_2 = 2x^2 \)[/tex].
- At [tex]\( x = 0 \)[/tex]:
[tex]\[ y_2 = 2 \cdot 0^2 = 0 \quad \text{and} \quad y_3 = 2^0 = 1 \][/tex]
Here, [tex]\( y_3 = 1 \)[/tex] when [tex]\( y_2 = 0 \)[/tex].
- Increase [tex]\( x \)[/tex] incrementally, we reach [tex]\( x = 1 \)[/tex]:
[tex]\[ y_2 = 2 \cdot 1^2 = 2 \quad \text{and} \quad y_3 = 2^1 = 2 \][/tex]
Here, [tex]\( y_3 = 2 \)[/tex], matching [tex]\( y_2 \)[/tex].
As the values of [tex]\( x \)[/tex] increase further, the exponential function [tex]\( y_3 \)[/tex] starts to surpass the quadratic function [tex]\( y_2 \)[/tex]. Therefore, the [tex]\( y \)[/tex]-value at which [tex]\( y_3 \)[/tex] surpasses [tex]\( y_2 \)[/tex] immediately after this point is:
[tex]\[ \boxed{1} \][/tex]
### Identifying the Fastest Growing Function
Among the three functions [tex]\( y_1 = 2x \)[/tex], [tex]\( y_2 = 2x^2 \)[/tex], and [tex]\( y_3 = 2^x \)[/tex], the exponential function grows the fastest. This is because exponential growth increases more rapidly than both linear and quadratic growth as [tex]\( x \)[/tex] increases.
Thus, the function that grows the fastest is:
[tex]\[ \boxed{\text{the exponential function}} \][/tex]
In summary:
- The exponential function [tex]\( y_3 \)[/tex] surpasses the linear function [tex]\( y_1 \)[/tex] at [tex]\( y = 1 \)[/tex].
- The exponential function [tex]\( y_3 \)[/tex] surpasses the quadratic function [tex]\( y_2 \)[/tex] at [tex]\( y = 1 \)[/tex].
- The exponential function grows the fastest.
