Which exponential function has an initial value of 2? f(x) = 2(3x) f(x) = 3(2x)

Which exponential function has an initial value of 2? f(x) = 2(3x) f(x) = 3(2x)

The Correct Answer and Explanation is:

ChatGPT said:

Correct Answer:
The exponential function with an initial value of 2 is:
f(x) = 2(3^x)


Explanation:

In an exponential function, the general form is:

f(x) = a(b^x)

Where:

  • a is the initial value (also called the y-intercept),
  • b is the base or growth/decay factor,
  • x is the exponent (usually the independent variable).

The initial value is the output of the function when x = 0. To determine the initial value of a function, substitute x = 0 into the equation:

  1. f(x) = 2(3^x)
    f(0) = 2(3^0) = 2(1) = 2
  2. f(x) = 3(2^x)
    f(0) = 3(2^0) = 3(1) = 3

The first function, f(x) = 2(3^x), has an initial value of 2, which matches the condition in the question. This makes it the correct exponential function.

The value a in the expression a(b^x) controls the vertical stretch or compression of the function and gives the value of the function at x = 0. The base b determines the rate of exponential growth (if b > 1) or decay (if 0 < b < 1).

In the case of f(x) = 2(3^x), the function starts at 2 when x is 0 and grows rapidly as x increases because the base 3 is greater than 1. This type of function models real-world phenomena such as compound interest, population growth, and bacterial reproduction, where the quantity increases by a fixed multiple over equal time intervals.

Understanding how the parameters of an exponential function affect its behavior is essential in interpreting and modeling exponential change in various applications.

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