Answer :
Let's analyze the function [tex]\( f(x) = 3(2.5)^x \)[/tex] in detail to determine its properties.
1. Is the function increasing?
- To determine if the function is increasing, we look at the derivative [tex]\( f'(x) \)[/tex]. A function is increasing if its derivative is greater than zero for all x. In the given situation, the function [tex]\( 3(2.5)^x \)[/tex] is not strictly increasing.
2. Initial Value:
- To find the initial value, we simply evaluate the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 3(2.5)^0 = 3 \times 1 = 3 \][/tex]
- So, the initial value of the function is [tex]\( 3 \)[/tex].
3. Limit as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex]:
- We consider the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive infinity. The expression [tex]\( (2.5)^x \)[/tex] grows without bound as [tex]\( x \)[/tex] increases. Hence:
[tex]\[ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} 3(2.5)^x = +\infty \][/tex]
- Therefore, the limit of the function as [tex]\( x \)[/tex] approaches positive infinity is [tex]\( \infty \)[/tex].
4. Limit as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
- For the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches negative infinity, note that [tex]\( (2.5)^x \)[/tex] tends towards zero since any base greater than one raised to a sufficiently large negative power approaches zero. Thus:
[tex]\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} 3(2.5)^x = 3 \times 0 = 0 \][/tex]
- Therefore, the limit of the function as [tex]\( x \)[/tex] approaches negative infinity is [tex]\( 0 \)[/tex].
Given these results:
1. The function is not increasing.
2. The initial value of the function [tex]\( f(0) \)[/tex] is [tex]\( 3 \)[/tex].
3. The limit of the function as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] is [tex]\( +\infty \)[/tex].
4. The limit of the function as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] is [tex]\( 0 \)[/tex].
These results match the true statements about the function [tex]\( f(x) = 3(2.5)^x \)[/tex].
1. Is the function increasing?
- To determine if the function is increasing, we look at the derivative [tex]\( f'(x) \)[/tex]. A function is increasing if its derivative is greater than zero for all x. In the given situation, the function [tex]\( 3(2.5)^x \)[/tex] is not strictly increasing.
2. Initial Value:
- To find the initial value, we simply evaluate the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 3(2.5)^0 = 3 \times 1 = 3 \][/tex]
- So, the initial value of the function is [tex]\( 3 \)[/tex].
3. Limit as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex]:
- We consider the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive infinity. The expression [tex]\( (2.5)^x \)[/tex] grows without bound as [tex]\( x \)[/tex] increases. Hence:
[tex]\[ \lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} 3(2.5)^x = +\infty \][/tex]
- Therefore, the limit of the function as [tex]\( x \)[/tex] approaches positive infinity is [tex]\( \infty \)[/tex].
4. Limit as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex]:
- For the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches negative infinity, note that [tex]\( (2.5)^x \)[/tex] tends towards zero since any base greater than one raised to a sufficiently large negative power approaches zero. Thus:
[tex]\[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} 3(2.5)^x = 3 \times 0 = 0 \][/tex]
- Therefore, the limit of the function as [tex]\( x \)[/tex] approaches negative infinity is [tex]\( 0 \)[/tex].
Given these results:
1. The function is not increasing.
2. The initial value of the function [tex]\( f(0) \)[/tex] is [tex]\( 3 \)[/tex].
3. The limit of the function as [tex]\( x \)[/tex] approaches [tex]\( +\infty \)[/tex] is [tex]\( +\infty \)[/tex].
4. The limit of the function as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] is [tex]\( 0 \)[/tex].
These results match the true statements about the function [tex]\( f(x) = 3(2.5)^x \)[/tex].