LailaNicole
LailaNicole LailaNicole
  • 10-08-2021
  • Mathematics
contestada

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

Respuesta :

reinhard10158
reinhard10158 reinhard10158
  • 10-08-2021

Answer:

f(0) = 0.25 = 1/4

f(x) = f(x-1) × 2/3

Step-by-step 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 ?

[tex]a - 1 = 0.25 \times {(2 \div 3)}^{ - 1} [/tex]

so, yes, a0 = a-1 × (2/3)

and a-1 = a-2 × (2/3)

...

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