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To compare the two functions, we need to examine their average rates of change and understand their behavior as [tex]\( x \)[/tex] increases.
Let’s first find the average rate of change for each function over a specific interval. We will compare the rates of change over different intervals.
For the function [tex]\( f(x) = 4x + 3 \)[/tex]:
The rate of change is constant because it is a linear function. The slope (rate of change) of [tex]\( f \)[/tex] is 4.
For the function [tex]\( g(x) = \left(\frac{6}{3}\right)^x = 2^x \)[/tex]:
The average rate of change over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \frac{g(b) - g(a)}{b - a} = \frac{2^b - 2^a}{b - a} \][/tex]
Now let's compare these over the interval [tex]\([1, 2]\)[/tex]:
For [tex]\( f(x) = 4x + 3 \)[/tex], the average rate of change is always 4.
For [tex]\( g(x) = 2^x \)[/tex]:
- When [tex]\( x = 1 \)[/tex], [tex]\( g(1) = 2^1 = 2 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( g(2) = 2^2 = 4 \)[/tex]
The average rate of change over [tex]\([1, 2]\)[/tex] is:
[tex]\[ \frac{g(2) - g(1)}{2 - 1} = \frac{4 - 2}{2 - 1} = 2 \][/tex]
Since 2 (the average rate of change of [tex]\( g \)[/tex] over [tex]\([1, 2]\)[/tex]) is less than 4 (the rate of change of [tex]\( f \)[/tex]) over the same interval.
However, as [tex]\( x \)[/tex] increases, the exponential function's rate of change grows significantly faster.
For a larger interval, say [tex]\([2, 3]\)[/tex]:
- When [tex]\( x = 2 \)[/tex], [tex]\( g(2) = 4 \)[/tex]
- When [tex]\( x = 3 \)[/tex], [tex]\( g(3) = 2^3 = 8 \)[/tex]
The average rate of change is:
[tex]\[ \frac{g(3) - g(2)}{3 - 2} = \frac{8 - 4}{3 - 2} = 4 \][/tex]
Here, the rate of change equals the rate of change of [tex]\( f \)[/tex].
For even larger intervals, the average rate of change of [tex]\( g \)[/tex] will eventually exceed 4.
Over the interval [tex]\([3, 4]\)[/tex] for instance:
- When [tex]\( x = 3 \)[/tex], [tex]\( g(3) = 8 \)[/tex]
- When [tex]\( x = 4 \)[/tex], [tex]\( g(4) = 2^4 = 16 \)[/tex]
The average rate of change becomes:
[tex]\[ \frac{g(4) - g(3)}{4 - 3} = \frac{16 - 8}{4 - 3} = 8 \][/tex]
This average rate of change of [tex]\( g \)[/tex] is greater than that of [tex]\( f \)[/tex].
Now, we can fill in the blanks:
Over the interval [tex]\(\boxed{[3, 4]}\)[/tex], the average rate of change of [tex]\( g \)[/tex] is greater than the average rate of change of [tex]\( f \)[/tex].
As the value of [tex]\( x \)[/tex] increases, the average rates of change of [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are [tex]\(\boxed{\text{constant} \text{ and } \text{increasing} } \)[/tex], respectively. When the value of [tex]\( x \)[/tex] is equal to 7, the value of [tex]\( \boxed{g(7) \text{ is } 128 } \)[/tex].
It can be further generalized that a quantity increasing exponentially will [tex]\(\boxed{eventually}\)[/tex] exceed a quantity increasing linearly.
Let’s first find the average rate of change for each function over a specific interval. We will compare the rates of change over different intervals.
For the function [tex]\( f(x) = 4x + 3 \)[/tex]:
The rate of change is constant because it is a linear function. The slope (rate of change) of [tex]\( f \)[/tex] is 4.
For the function [tex]\( g(x) = \left(\frac{6}{3}\right)^x = 2^x \)[/tex]:
The average rate of change over an interval [tex]\([a, b]\)[/tex] is given by:
[tex]\[ \frac{g(b) - g(a)}{b - a} = \frac{2^b - 2^a}{b - a} \][/tex]
Now let's compare these over the interval [tex]\([1, 2]\)[/tex]:
For [tex]\( f(x) = 4x + 3 \)[/tex], the average rate of change is always 4.
For [tex]\( g(x) = 2^x \)[/tex]:
- When [tex]\( x = 1 \)[/tex], [tex]\( g(1) = 2^1 = 2 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( g(2) = 2^2 = 4 \)[/tex]
The average rate of change over [tex]\([1, 2]\)[/tex] is:
[tex]\[ \frac{g(2) - g(1)}{2 - 1} = \frac{4 - 2}{2 - 1} = 2 \][/tex]
Since 2 (the average rate of change of [tex]\( g \)[/tex] over [tex]\([1, 2]\)[/tex]) is less than 4 (the rate of change of [tex]\( f \)[/tex]) over the same interval.
However, as [tex]\( x \)[/tex] increases, the exponential function's rate of change grows significantly faster.
For a larger interval, say [tex]\([2, 3]\)[/tex]:
- When [tex]\( x = 2 \)[/tex], [tex]\( g(2) = 4 \)[/tex]
- When [tex]\( x = 3 \)[/tex], [tex]\( g(3) = 2^3 = 8 \)[/tex]
The average rate of change is:
[tex]\[ \frac{g(3) - g(2)}{3 - 2} = \frac{8 - 4}{3 - 2} = 4 \][/tex]
Here, the rate of change equals the rate of change of [tex]\( f \)[/tex].
For even larger intervals, the average rate of change of [tex]\( g \)[/tex] will eventually exceed 4.
Over the interval [tex]\([3, 4]\)[/tex] for instance:
- When [tex]\( x = 3 \)[/tex], [tex]\( g(3) = 8 \)[/tex]
- When [tex]\( x = 4 \)[/tex], [tex]\( g(4) = 2^4 = 16 \)[/tex]
The average rate of change becomes:
[tex]\[ \frac{g(4) - g(3)}{4 - 3} = \frac{16 - 8}{4 - 3} = 8 \][/tex]
This average rate of change of [tex]\( g \)[/tex] is greater than that of [tex]\( f \)[/tex].
Now, we can fill in the blanks:
Over the interval [tex]\(\boxed{[3, 4]}\)[/tex], the average rate of change of [tex]\( g \)[/tex] is greater than the average rate of change of [tex]\( f \)[/tex].
As the value of [tex]\( x \)[/tex] increases, the average rates of change of [tex]\( f \)[/tex] and [tex]\( g \)[/tex] are [tex]\(\boxed{\text{constant} \text{ and } \text{increasing} } \)[/tex], respectively. When the value of [tex]\( x \)[/tex] is equal to 7, the value of [tex]\( \boxed{g(7) \text{ is } 128 } \)[/tex].
It can be further generalized that a quantity increasing exponentially will [tex]\(\boxed{eventually}\)[/tex] exceed a quantity increasing linearly.
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