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Let's analyze how modifying the function [tex]\( f(x) = a b^x \)[/tex] by increasing the value of [tex]\( a \)[/tex] by 2 affects its domain and range.
### Domain Analysis
The original function [tex]\( f(x) = a b^x \)[/tex] is an exponential function. For exponential functions, the independent variable [tex]\( x \)[/tex] can take any real number value. This means the domain of the function [tex]\( f(x) \)[/tex] is all real numbers, denoted as [tex]\( (-\infty, \infty) \)[/tex].
When we increase the value of [tex]\( a \)[/tex] by 2, the new function becomes [tex]\( f_{new}(x) = (a+2) b^x \)[/tex].
- Since the operation modifies only the coefficient [tex]\( a \)[/tex] and does not affect the exponent or the base [tex]\( b \)[/tex], the domain of the new function remains the same. That is, the new domain is still all real numbers, [tex]\( (-\infty, \infty) \)[/tex].
Thus, the statement "The domain stays the same" is True.
- There is no change to the domain that would restrict [tex]\( x \)[/tex] to values greater than 2 or greater than or equal to 2, so the statements "The domain becomes [tex]\( x > 2 \)[/tex]" and "The domain becomes [tex]\( x \geq 2 \)[/tex]" are both False.
### Range Analysis
Now let's consider the range of the function.
For the original function [tex]\( f(x) = a b^x \)[/tex]:
- If [tex]\( a > 0 \)[/tex] and [tex]\( b > 1 \)[/tex], the range of [tex]\( f(x) \)[/tex] is [tex]\( (0, \infty) \)[/tex] because [tex]\( b^x \)[/tex] is always positive and so is [tex]\( a \)[/tex].
- If [tex]\( a < 0 \)[/tex], the range would be [tex]\( (-\infty, 0) \)[/tex].
After increasing [tex]\( a \)[/tex] by 2, the new function is [tex]\( f_{new}(x) = (a+2) b^x \)[/tex].
- If the original [tex]\(a > 0\)[/tex]: The range of the original function was [tex]\( (0, \infty) \)[/tex]. With [tex]\(a+2\)[/tex], the range of the new function is now [tex]\( (2, \infty) \)[/tex].
- If the original [tex]\(a < 0\)[/tex]: The range of the original function was [tex]\( (-\infty, 0) \)[/tex]. With [tex]\(a+2\)[/tex], the range of the new function could potentially still be negative or include non-negative values, depending on [tex]\(a\)[/tex].
However, in a common scenario where [tex]\(a > 0\)[/tex], the range of the new function shifts upwards by 2 units.
Given the common case:
- The statement "The range stays the same" is False because the range shifts upwards.
- The statement "The range becomes [tex]\( y>2 \)[/tex]" is True because the entire range moves up by 2 units.
- The statement "The range becomes [tex]\( y \geq 2 \)[/tex]" is False because the range [tex]\( y > 2 \)[/tex] excludes 2 itself as [tex]\( f(x) \)[/tex] will never equal exactly 2.
In summary:
- The domain stays the same: True.
- The range stays the same: False.
- The range becomes [tex]\( y > 2 \)[/tex]: True.
- The domain becomes [tex]\( x > 2 \)[/tex]: False.
- The range becomes [tex]\( y \geq 2 \)[/tex]: False.
- The domain becomes [tex]\( x \geq 2 \)[/tex]: False.
Explicitly, the correct selections are:
- The range becomes [tex]\( y > 2 \)[/tex].
- The domain stays the same.
### Domain Analysis
The original function [tex]\( f(x) = a b^x \)[/tex] is an exponential function. For exponential functions, the independent variable [tex]\( x \)[/tex] can take any real number value. This means the domain of the function [tex]\( f(x) \)[/tex] is all real numbers, denoted as [tex]\( (-\infty, \infty) \)[/tex].
When we increase the value of [tex]\( a \)[/tex] by 2, the new function becomes [tex]\( f_{new}(x) = (a+2) b^x \)[/tex].
- Since the operation modifies only the coefficient [tex]\( a \)[/tex] and does not affect the exponent or the base [tex]\( b \)[/tex], the domain of the new function remains the same. That is, the new domain is still all real numbers, [tex]\( (-\infty, \infty) \)[/tex].
Thus, the statement "The domain stays the same" is True.
- There is no change to the domain that would restrict [tex]\( x \)[/tex] to values greater than 2 or greater than or equal to 2, so the statements "The domain becomes [tex]\( x > 2 \)[/tex]" and "The domain becomes [tex]\( x \geq 2 \)[/tex]" are both False.
### Range Analysis
Now let's consider the range of the function.
For the original function [tex]\( f(x) = a b^x \)[/tex]:
- If [tex]\( a > 0 \)[/tex] and [tex]\( b > 1 \)[/tex], the range of [tex]\( f(x) \)[/tex] is [tex]\( (0, \infty) \)[/tex] because [tex]\( b^x \)[/tex] is always positive and so is [tex]\( a \)[/tex].
- If [tex]\( a < 0 \)[/tex], the range would be [tex]\( (-\infty, 0) \)[/tex].
After increasing [tex]\( a \)[/tex] by 2, the new function is [tex]\( f_{new}(x) = (a+2) b^x \)[/tex].
- If the original [tex]\(a > 0\)[/tex]: The range of the original function was [tex]\( (0, \infty) \)[/tex]. With [tex]\(a+2\)[/tex], the range of the new function is now [tex]\( (2, \infty) \)[/tex].
- If the original [tex]\(a < 0\)[/tex]: The range of the original function was [tex]\( (-\infty, 0) \)[/tex]. With [tex]\(a+2\)[/tex], the range of the new function could potentially still be negative or include non-negative values, depending on [tex]\(a\)[/tex].
However, in a common scenario where [tex]\(a > 0\)[/tex], the range of the new function shifts upwards by 2 units.
Given the common case:
- The statement "The range stays the same" is False because the range shifts upwards.
- The statement "The range becomes [tex]\( y>2 \)[/tex]" is True because the entire range moves up by 2 units.
- The statement "The range becomes [tex]\( y \geq 2 \)[/tex]" is False because the range [tex]\( y > 2 \)[/tex] excludes 2 itself as [tex]\( f(x) \)[/tex] will never equal exactly 2.
In summary:
- The domain stays the same: True.
- The range stays the same: False.
- The range becomes [tex]\( y > 2 \)[/tex]: True.
- The domain becomes [tex]\( x > 2 \)[/tex]: False.
- The range becomes [tex]\( y \geq 2 \)[/tex]: False.
- The domain becomes [tex]\( x \geq 2 \)[/tex]: False.
Explicitly, the correct selections are:
- The range becomes [tex]\( y > 2 \)[/tex].
- The domain stays the same.
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