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

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.

answer: i wpould go with c

step-by-step explanation: