Question

Which is the graph of the sequence defined by the function f(x+1)=2/3f(x) if the initial value of the sequence is 108

Answer 1

First four terms of this sequence are:

$f_0=f(0)=108,\\f_1=f(1)=\dfrac{2}{3}f(0)= \dfrac{2}{3}\cdot 108=72,\\f_2=f(2)=\dfrac{2}{3}f(1)= \dfrac{2}{3}\cdot 72=48,\\ f_3=f(3)=\dfrac{2}{3}f(2)= \dfrac{2}{3}\cdot 48=32$.

This sequence is decreasing and seems to be exponential. See whether you can determine such a and b, that $f(x)=a\cdot b^x$:

1.

$f(0)=a\cdot b^0=a,\\ f(0)=108$, then a=108,

2.

$f(1)=108\cdot b^1=108 b,\\ f(1)=72$, then $b=\dfrac{72}{108} =\dfrac{2}{3}$.

3. Check for x=2 and x=3:

a) x=2:

$f(2)=108\cdot \left(\dfrac{2}{3}\right)^2=108\cdot \dfrac{4}{9} =48$ (true),

b) x=3:

$f(3)=108\cdot \left(\dfrac{2}{3}\right)^3=108\cdot \dfrac{8}{27} =32$ (true).

4. Check for all x:

$f(x+1)=108\cdot \left(\dfrac{2}{3}\right)^{x+1},\\ f(x)=108\cdot \left(\dfrac{2}{3}\right)^x$.

Equate f(x+1) and 2/3f(x):

$108\cdot \left(\dfrac{2}{3}\right)^{x+1}=\dfrac{2}{3}\cdot 108\cdot \left(\dfrac{2}{3}\right)^x ,\\\left(\dfrac{2}{3}\right)^{x+1}=\left(\dfrac{2}{3}\right)^{x+1}$ is true equality for all x.

Answer: the graph is the graph of the function $f(x)=108\cdot \left(\dfrac{2}{3}\right)^x$.

Answer 2

Answer:

See picture attached.

Step-by-step explanation:

For x = 1, f(x) = 108. For the next x-value which is 2, that is, x+1, f(x+1) = (2/3)*f(x) = (2/3)*108 = 72.   For the next x-value, which is 3, f(x+1) = (2/3)*f(x) = (2/3)*72 = 48 . For the next x-value, which is 4, f(x+1) = (2/3)*f(x) = (2/3)*48 = 32 . And so on.

x f(x)

1 108

2 72

3 48

4 32