Question

Write a recrusive rule for the exponential function
f(x)=0.25(2/3)^x​

Answer 1

f(0) = 0.25 = 1/4

f(x) = f(x-1) × 2/3

Explanation:

a0 = 0.25×(2/3)⁰ = 0.25×1 = 0.25 = 1/4

a1 = 0.25×(2/3)¹ = 0.25×(2/3) = 1/4 × 2/3 = 2/12 = 1/6 =

= a0 × 2/3

a2 = 0.25×(2/3)² = 0.25×(4/9) = 1/4 × 4/9 = 4/36 = 1/9 =

= a1 × 2/3

=>

an = an-1 × 2/3, n >= 1

but what about negative x ?

is a0 = a-1 × 2/3 ?

`a - 1 = 0.25 * {(2 / 3)}^( - 1)`

so, yes, a0 = a-1 × (2/3)

and a-1 = a-2 × (2/3)

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