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What is the initial value in the equation f(x) = 350(1 - 0.12)^x? *

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Given:

The function is:


f(x)=350(1-0.12)^x

To find:

The initial value of the function.

Solution:

We have,


f(x)=350(1-0.12)^x

The value of the function at x=0 is called the initial value of the function.

For x=0, we get


f(0)=350(1-0.12)^0


f(0)=350(0.88)^0

Clearly 0.88 is a non zero number and zero to the power of a non zero number is always 1.


f(0)=350(1)


f(0)=350

Therefore, the initial value of the function is 350.

User Tasawer Nawaz
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