Let $f(x) = x^2 - 2x$. Find all real numbers $x$ such that $f(x) = f(f(x))$. List your solutions separated by commas in any order​

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Explanation:

The given function is f(x) = x^2 - 2x

Let's apply function composition like so

In the second step, I replaced every x with f(x). Then in the third step, I replaced f(x) with x^2-2x on the right side.

Setting this equal to f(x) gets us

which is a bit cluttered. However, we have "x^2-2x" show up three times. Let's say w = x^2-2x

If we replaced all those "x^2-2x" expressions with w, then we get,

Which looks more manageable

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Let's solve for w

w^2 - 2w = w

w^2 - 2w-w = 0

w^2 - 3w = 0

w(w - 3) = 0

w = 0 or w-3 = 0

w = 0 or w = 3

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If w = 0, then,

w = x^2 - 2x

x^2 - 2x = w

x^2 - 2x = 0

x(x - 2) = 0

x = 0 or x-2 = 0

x = 0 or x = 2 are two solutions (out of four total)

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Or, if w = 3, then,

w = x^2 - 2x

x^2 - 2x = w

x^2 - 2x = 3

x^2 - 2x - 3 = 0

(x - 3)(x + 1) = 0

x-3 = 0 or x+1 = 0

x = 3 or x = -1 are the other two solutions

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To summarize, the four solutions are:  x = -1, x = 0, x = 2, x = 3

answer: sorry i can't you.

step-by-step explanation:

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).