Compare the functions below. select the true statements. over the interval [2, 3], the average rate of change of g is lower than that of both f and h. as x increases on the interval [0, ∞), the rate of change of g eventually exceeds the rate of change of both f and h. when x ≈ 4, the value of f(x) exceeds the values of both g(x) and h(x). as x increases on the interval [0, ∞), the rate of change of f eventually exceeds the rate of change of both g and h. a quantity increasing exponentially eventually exceeds a quantity growing quadratically or linearly. when x ≈ 8, the value of h(x) exceeds the values of both f(x) and g(x). plese need asap !

The true statements are:

4) As x increases on the interval [0, ∞), the rate of change of f eventually exceeds the rate of change of both g and h.

5) A quantity increasing exponentially eventually exceeds a quantity growing quadratically or linearly.

Step-by-step explanation:

f(x)=3^x+2

g(x)=20x+4

h(x)=2x^2+5x+2

1) Over the interval [2, 3], the average rate of change of g is lower than that of both f and h.

Over the interval [a,b], the average rate of change of a function "j" is:

rj=[j(b)-j(a)]/(b-a); with a=2 and b=3

rj=[j(3)-j(2)]/(3-2)

rj=[j(3)-j(2)]/(1)

rj=j(3)-j(2)

For g(x):

rg=g(3)-g(2)

g(3)=20(3)+4→g(3)=60+4→g(3)=64

g(2)=20(2)+4→g(2)=40+4→g(2)=44

rg=64-44→rg=20

For f(x):

rf=f(3)-f(2)

f(3)=3^3+2→f(3)=27+2→f(3)=29

f(2)=3^2+2→f(2)=9+2→f(2)=11

rf=29-11→rf=18

For h(x):

rh=h(3)-h(2)

h(3)=2(3)^2+5(3)+2→h(3)=2(9)+15+2→h(3)=18+15+2→h(3)=35

h(2)=2(2)^2+5(2)+2→h(2)=2(4)+10+2→h(2)=8+10+2→h(2)=20

rh=35-20→rh=15

Over the interval [2, 3], the average rate of change of g (20) is greater than that of both f (18) and h (15), then the first statement is false.

2) As x increases on the interval [0, ∞), the rate of change of g eventually exceeds the rate of change of both f and h.

False, because of f(x) is an exponential function, the rate of f eventually exceeds the rate of change of both g and h.

3) When x=4, the value of f(x) exceeds the values of both g(x) and h(x).

x=4→f(4)=3^4+2=81+2→f(4)=83

x=4→g(4)=20(4)+4=80+4→g(4)=84

x=4→h(4)=2(4)^2+5(4)+2=2(16)+20+2=32+20+2→h(4)=54

When x=4, the value of f(x) (83) exceeds only the value of h(x) (54), then the third statement is false.

4) As x increases on the interval [0, ∞), the rate of change of f eventually exceeds the rate of change of both g and h.

True, because of f(x) is an exponential function.

5) A quantity increasing exponentially eventually exceeds a quantity growing quadratically or linearly.

True.

6) When x=8, the value of h(x) exceeds the values of both f(x) and g(x).

x=8→f(8)=3^8+2=6,561+2→f(8)=6,563

x=8→g(8)=20(8)+4=160+4→g(8)=164

x=8→h(8)=2(8)^2+5(8)+2=2(64)+40+2=128+40+2→h(8)=170

When x=8, the value of h(x) (170) exceeds only the value of g(x) (164), then the sixth statement is false.

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).