A diver jumps up and over from a cliff in order to avoid rocks and land in the deeper water. The path of the dive can be modeled using the function , where f(x) is the height, in feet, of the diver at x time in seconds. What is the maximum height, in feet, the diver will reach? Round your answer to the nearest hundredth if necessary. Do not include units in your answer.

Find the maximum height by solving for when the derivative of f(x) = 0 and the sign changes from positive to negative:

f(x) = -16x² + 2x + 48

Find the derivative of each term:

f'(x) = -32x + 2

Set the equation equal to 0:

0 = -32x + 2

-2 = -32x

x = 1/16 sec

Validate that the derivative changes from + to - at this value by plugging in values into the derivative formula on both sides of 1/16:

f'(0) = 2, +

f'(2) = -62, -

Therefore, there is a maximum at x = 1/16 because the function changes from positive to negative. Find the height at x = 16 seconds. (Plug it into the original equation)

f(1/16) = -16(1/16)² + 2(1/16) + 48

f(1/16) = -1/16 + 2/16 + 48

f(1/16) = 1/16 + 48

f(1/16) = 48 1/16 or 48.0625 ft

Round to nearest hundredth: ≈48.06 feet.

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).