Explain how you can approximate a cube root when the cube is not a perfect cube.

Step-by-step

You can  use Newton's method as follows:

It is an iterative method, meaning that from an approximation (i.e. an approximate solution), we can refine the answer to get a closer approximation.  By repeating the process (iteration), we can get an answer as accurate as we wish.

The basis of the formula is based on

x1=x0-f(x0)/f'(x0)

where

x0 is a given approximation

x1 is a better approximation

f(x) is a given function (for which we need the roots)

f'(x) is the derivative of the function.

Skipping all details and applying directly to find the cube root, we have

N=the number for which the cube root is desired

f(x)=x^3-N

f'(x)=3x^2

and x0 is an initial approximation that we need to provide (from the integer cubes, for example).

Say we need to find the cube-root of 124.

We know that 5^3 = 125, a rather close approximation, but the cube root is slightly less than 5.

To find a better approximation, we apply Newton's method, and calculate mentally:

x0=5, N=124

x1 = x0-(x0^3-124)/(3(x0^2))    [next substitute values]

= 5 - (125-124)/(3(5^2))     [ next simplify ]

= 5-1/75    [next, rearrange to calculate mentally ]

=5-(1/25)/3  [ next, substitute 1/25 = 0.04, 1/3 division can be done mentally ]

= 5 - 0.04/3    [ divide 0.04/3 mentally ]

= 5 - 0.013333    [ subtract ]

= 4.98667

(exact value = 4.98663, first approximation is already quite close)

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).

35 is an outlier

step-by-step explanation: