Question

Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. In your final answer, include the table of function values.
f(x) = 2x
h(x) = x^3 + x + 8

Answer 1

F(x)=2ˣ
f(x)
(x,y)
(0,1)
(1,2)
(2,4)
(3,8)
(4,16)
(5,32)
(6,64)
(7,128)
(8,256)
(9,512)
(10,1024)

h(x)
(x,y)
(0,8)
(1,10)
(2,18)
(3,38)
(4,76)
(5,138)
(6,230)
(7,358)
(8,528)
(9,736)
(10,1018)

so somwehre between x=9 and x=10

hmm
f(9.5)=724.007
h(9.5)=874.875

go to 9.75
f(9.75)=944.609
h(9.75)=861.078

up

9.9
f(9.9)=955.426
h(9.9)=988.199
up

9.95
f(9.95)=989.119
h(9.95)=1003.02
up

9.99
f(9.99)=1016.93
h(9.99)=1014.99

a bit down
9.98
f(9.98)=1009.9
h(9.98)=1011.99

a bit up
9.985
f(9.985)=1013.41
h(9.985)=1013.49
pretty close, ya

so about at 9.985=x
if we graphed, we get that x=9.98512070914982..., so we are pretty close

Answer 2

f(x) = 2^x; h(x) = x^3 + x + 8

Table

x      f(x) =  2^x             h(x) = x^3 + x + 8

0      2^0 = 1                0  + 0 + 8 = 8

1     2^1 = 2                 1^3 + 1 + 8 = 10

2      2^2 = 4                2^3 + 2 + 8 = 8 + 2 + 8 = 18

3      2^3 = 8                3^3  + 3 + 8 = 27 + 3 + 8 = 38

4      2^2 = 16              4^3 + 4 + 8 = 76

10    2^10 = 1024       10^3  +10 + 8 = 1018

 9      2^9 =   512        9^3 + 9 + 8 = 729 + 9 + 8 = 746

Answer: an approximate value of 10