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Sagot :
To compare the two functions [tex]\( f(x) = \frac{1}{50} \cdot 3^x \)[/tex] and [tex]\( g(x) = \frac{1}{5} \cdot x^2 \)[/tex] in terms of their rates of change, we need to find their derivatives. The derivative of a function gives us the rate at which the function changes with respect to [tex]\( x \)[/tex].
1. Finding the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{1}{50} \cdot 3^x \][/tex]
The derivative of an exponential function [tex]\( a \cdot b^x \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{50} \cdot 3^x \right) = \frac{1}{50} \cdot 3^x \cdot \ln(3) \][/tex]
From the simplified numerical result, this is approximately:
[tex]\[ f'(x) \approx 0.0219722457733623 \cdot 3^x \][/tex]
2. Finding the derivative of [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \frac{1}{5} \cdot x^2 \][/tex]
The derivative of a power function [tex]\( a \cdot x^n \)[/tex] is given by:
[tex]\[ g'(x) = \frac{d}{dx} \left( \frac{1}{5} \cdot x^2 \right) = \frac{1}{5} \cdot 2x = \frac{2x}{5} \][/tex]
From the simplified numerical result, this is:
[tex]\[ g'(x) = 0.4x \][/tex]
3. Comparing the derivatives:
To determine which rate of change exceeds the other as [tex]\( x \)[/tex] increases, we compare the obtained derivatives [tex]\( f'(x) \)[/tex] and [tex]\( g'(x) \)[/tex]:
[tex]\[ f'(x) = 0.0219722457733623 \cdot 3^x \][/tex]
[tex]\[ g'(x) = 0.4x \][/tex]
- For small values of [tex]\( x \)[/tex], [tex]\( 0.4x \)[/tex] might be greater than [tex]\( 0.0219722457733623 \cdot 3^x \)[/tex].
- However, because the exponential function [tex]\( 3^x \)[/tex] grows significantly faster than the linear function [tex]\( x \)[/tex] as [tex]\( x \)[/tex] becomes large, [tex]\( 0.0219722457733623 \cdot 3^x \)[/tex] will eventually exceed [tex]\( 0.4x \)[/tex].
Thus, as [tex]\( x \)[/tex] increases, the rate of change of [tex]\( f(x) \)[/tex] exceeds the rate of change of [tex]\( g(x) \)[/tex]. Therefore, the correct statement is:
B. As [tex]\( x \)[/tex] increases, the rate of change of [tex]\( f(x) \)[/tex] exceeds the rate of change of [tex]\( g(x) \)[/tex].
1. Finding the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{1}{50} \cdot 3^x \][/tex]
The derivative of an exponential function [tex]\( a \cdot b^x \)[/tex] is given by:
[tex]\[ f'(x) = \frac{d}{dx} \left( \frac{1}{50} \cdot 3^x \right) = \frac{1}{50} \cdot 3^x \cdot \ln(3) \][/tex]
From the simplified numerical result, this is approximately:
[tex]\[ f'(x) \approx 0.0219722457733623 \cdot 3^x \][/tex]
2. Finding the derivative of [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \frac{1}{5} \cdot x^2 \][/tex]
The derivative of a power function [tex]\( a \cdot x^n \)[/tex] is given by:
[tex]\[ g'(x) = \frac{d}{dx} \left( \frac{1}{5} \cdot x^2 \right) = \frac{1}{5} \cdot 2x = \frac{2x}{5} \][/tex]
From the simplified numerical result, this is:
[tex]\[ g'(x) = 0.4x \][/tex]
3. Comparing the derivatives:
To determine which rate of change exceeds the other as [tex]\( x \)[/tex] increases, we compare the obtained derivatives [tex]\( f'(x) \)[/tex] and [tex]\( g'(x) \)[/tex]:
[tex]\[ f'(x) = 0.0219722457733623 \cdot 3^x \][/tex]
[tex]\[ g'(x) = 0.4x \][/tex]
- For small values of [tex]\( x \)[/tex], [tex]\( 0.4x \)[/tex] might be greater than [tex]\( 0.0219722457733623 \cdot 3^x \)[/tex].
- However, because the exponential function [tex]\( 3^x \)[/tex] grows significantly faster than the linear function [tex]\( x \)[/tex] as [tex]\( x \)[/tex] becomes large, [tex]\( 0.0219722457733623 \cdot 3^x \)[/tex] will eventually exceed [tex]\( 0.4x \)[/tex].
Thus, as [tex]\( x \)[/tex] increases, the rate of change of [tex]\( f(x) \)[/tex] exceeds the rate of change of [tex]\( g(x) \)[/tex]. Therefore, the correct statement is:
B. As [tex]\( x \)[/tex] increases, the rate of change of [tex]\( f(x) \)[/tex] exceeds the rate of change of [tex]\( g(x) \)[/tex].
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