the value of an exponential decreasing function, f(x), is always greater than zero
Is the value of an exponential decreasing function, ( f(x) ), always greater than zero?
Answer: Let’s explore the concept of an exponential decreasing function. An exponential function generally takes the form ( f(x) = a \cdot b^x ), where:
- ( a ) is a constant, and ( a > 0 ).
- ( b ) is the base, and for a decreasing function, ( 0 < b < 1 ).
Properties of Exponential Functions:
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Base Greater Than Zero: For ( b > 0 ), the exponential function is always defined for all real numbers ( x ).
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Positive Outputs: When ( 0 < b < 1 ), each power of ( b ) decreases but remains positive, because any positive number raised to any real power is positive.
Why ( f(x) > 0 ):
- If ( a > 0 ) and ( b^x > 0 ), the product ( f(x) = a \cdot b^x ) will also be greater than zero.
- Negative or zero values of ( a ) could make the function zero or negative, but by definition, for exponential decay, ( a > 0 ).
Example:
Consider ( f(x) = 4 \cdot (0.5)^x ):
- Here, ( a = 4 ) and ( b = 0.5 ), so the function is always positive, ( f(x) > 0 ).
Summary: An exponential decreasing function, ( f(x) = a \cdot b^x ) with ( a > 0 ) and ( 0 < b < 1 ), will always have values greater than zero for all real ( x ). @anonymous4