The domain of startfraction f over g endfraction f g consists of numbers x for which g left parenthesis x right parenthesisg(x) 0 that are in the domains of?

The domain of f/g consists of numbers x for which g(x) cannot equal 0 that are in the domains of both f and g.

Let’s take this equation as an example:

If f(x) = 3x - 5 and g(x) = square root of x-5, what is the domain of (f/g)x. 

For x to be in the domain of (f/g)(x), it must be in the domain of f and in the domain of g since (f/g)(x) = f(x)/g(x). We also need to ensure that g(x) is not zero since f(x) is divided by g(x). Therefore, there are 3 conditions.

x must be in the domain of f: f(x) = 3x -5 are in the domain of x and all real numbers x.

x must be in the domain of g: g(x) = √(x - 5) so x - 5 ≥ 0 so x ≥ 5.

g(x) can not be 0: g(x) = √(x - 5) and √(x - 5) = 0 gives x = 5 so x ≠ 5.

Hence to x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x satisfying x > 5.

Thus, satisfying satisfy all three conditions, x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x satisfying x > 5.

b. x + 1

step-by-step explanation:

i'am smart

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).