Question

find the values of the exponential function f(x) when x=1 2 3 4 5
identify wether it is a growth or decay

1.f(x)=(1/3)^x-1

2.f(x) = -(1/2)*+1

3.f(x)=5(2)^x

plss help ‍♂​

Answer 1

3)

:f(x)=5(2)^x

f(1)=5(2)^1

=10

:f(2)=5(2)^2

=20

:f(3)=5(2)^3

=40

:f(4)= 5(2)^4

=5(16)

=80

:f(5)=5(2)^5

=5(32)

=160

Answer 2

Given:

The exponential function f(x)  

x=1 2 3 4 5

(1)  $f(x)=(\frac{1}{3} )^{x-1}$

• given that x = 1, 2, 3, 4, 5

• so first solve for x=1

$f(x)=(\frac{1}{3} )^{1-1}$

$f(1)=(\frac{1}{3} )^{0}\\f(1)=1$

for x = 2

$f(2)=(\frac{1}{3} )^{2-1}\\\\f(2)=(\frac{1}{3} )^{1}\\\\f(2)=\frac{1}{3}$

for  x=3

$f(3)=(\frac{1}{3} )^{3-1}\\\\f(3)=(\frac{1}{3} )^{2}\\\\f(3)=(\frac{1}{9} )$

For x= 4

$f(4)=(\frac{1}{3} )^{4-1}\\\\f(4)=(\frac{1}{3} )^{3}\\\\f(4)=(\frac{1}{27} )$

For x=5

$f(5)=(\frac{1}{3} )^{5-1}\\\\f(5)=(\frac{1}{3} )^{4}\\\\f(5)=(\frac{1}{81} )$

Answer is $f(x)= 1, \frac{1}{3}, \frac{1}{9} ,\frac{1}{27}, \frac{1}{81}$ its decay.

(2) $f(x)=-(\frac{1}{2} )^{x}+1$

For x =1

$f(1)=-(\frac{1}{2} )^{1}+1 \\\\f(1)=-\frac{1}{2} +1\\\\f(x)= \frac{1}{2}\\\\f(1)=0.5$

For x=2

$f(2)=-(\frac{1}{2} )^{2}+1 \\\f(2)=-(\frac{1}{4} )+1 \\\f(2) = 0.75$

For x=3

$f(3)=-(\frac{1}{2} )^{2}+1 \\\\f(3)=-(\frac{1}{8} )+1 \\\f(3) = 0.875$

For x=4

$f(4)=-(\frac{1}{2} )^{2}+1 \\\\f(4)=-(\frac{1}{16} )+1 \\\\f(4) = 0.93$

For x=5

$f(5)=-(\frac{1}{2} )^{2}+1 \\\\f(5)=-(\frac{1}{32} )+1 \\\f(5) = 0.96$

f(x) = 0.5, 0.75, 0.875, 0.93, 0.96 it is growth

(3) $f(x)=5(2)^{x}$

for x=1

$f(1)=5(2)^{1}\\\f(1)=10$

for x= 2

$f(2)=5(2)^{2}\\\f(2)=20$

for x=3

$f(3)=5(2)^{3}\\\f(3)=40$

for x=4

$f(4)=5(2)^{4}\\\f(4)=480$

for x=5

$f(5)=5(2)^{5}\\\f(5)=160$

f(x)=10, 20, 40, 80, 160 it is growth

Hence the answer is

(1) 1, 1/3, 1/9, 1/27, 1/81 its decay

(2) 0.5, 0.75, 0.875, 0.93, 0.96 its growth

(3) 10, 20, 40, 80, 160 its growth