Question

Someone do the work pls

Answer 1

`f(2) = 12`

`f(5) =320`

`f(-2) =(5)/(2)`

`g(3) = (1)/(135)`

Explanation:

Solving (1):

`f(x) = (1)/(3)*6^x` at
`x=2`

Substitute 2 for x in f(x)

`f(2) = (1)/(3)*6^2`

`f(2) = (1)/(3)*36`

`f(2) = (1*36)/(3)`

`f(2) = (36)/(3)`

`f(2) = 12`

Solving (2):

`f(n) =10*2^n` at
`n =5`

Substitute 5 for n in f(n)

`f(5) =10*2^5`

`f(5) =10*32`

`f(5) =320`

Solving (3):

`f(n) =10*2^n` at
`n =-2`

Substitute -2 for n in f(n)

`f(-2) =10*2^{-2`

`f(-2) =10*(1)/(2^2)`

`f(-2) =10*(1)/(4)`

`f(-2) =(10*1)/(4)`

`f(-2) =(10)/(4)`

`f(-2) =(5)/(2)`

Solving (4):

`g(x) = (1)/(5)*((1)/(3))^x` at
`x = 3`

Substitute 3 for x in g(x)

`g(3) = (1)/(5)*((1)/(3))^3`

`g(3) = (1)/(5)*((1^3)/(3^3))`

`g(3) = (1)/(5)*(1)/(27)`

`g(3) = (1)/(135)`

Solving (5):

Graph of
`f(x) = 4 * 2^x`

First, we determine the points to plot the graph.

When
`x = 0;\ f(0) = 4 * 2^0 = 4*1 = 4`

When
`x = 2;\ f(2) = 4 * 2^2 = 4*2 = 16`

When
`x = 4;\ f(4) = 4 * 2^4 = 4*16 = 64`

So, we have: (0,4), (2,16) and (4,16)

See attachment 1 for graph

Solving (6):

Graph of
`f(x) = 4 * ((1)/(2))^x`

First, we determine the points to plot the graph.

When
`x = 0;\ f(0) = 4 * ((1)/(2))^0 = 4*1 = 4`

When
`x = 2;\ f(2) = 4 * ((1)/(2))^2 = 4*(1)/(4) = 1`

When
`x = 4;\ f(4) = 4 * ((1)/(2))^4 = 4*(1)/(16) = (1)/(4)`

So, we have: (0,4), (2,1) and (4,1/4)

See attachment 2 for graph