# Given an exponential function for compounding interest, A(x) = P(.91)^x, what is the rate of change?

## Question:

Given an exponential function for compounding interest, {eq}A(x) = P(.91)^x {/eq}, what is the rate of change?

## Growth/Decay:

The exponential equation, {eq}A(x)=ab^x {/eq} is an exponenial equation where {eq}a {/eq} is the initial quantity. It

(i) represents the exponential growth if {eq}b>1 {/eq} and this case, the rate of change is, {eq}(b-1) \times 100 {/eq}.

(ii) represents the exponential decay if {eq}b<1 {/eq} and this case, the rate of change is, {eq}(1-b) \times 100 {/eq}.

## Answer and Explanation:

The given function is:

$$A(x) = P(.91)^x $$

Comparing this with {eq}A(x)= a b^x {/eq}, we get {eq}b=0.91 {/eq}.

Here {eq}b=0.91 <1 {/eq}.

So the function represents the exponential decay and the rate of change in this case is,

$$(1-b) \times 100 = (1-0.91) \times 100 = 0.09 \times 100 = \boxed{\mathbf{9 \%}} $$

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