A portion of the graph of y = f(x) is shown in red below, where f(x) is a quadratic function. The distance between grid lines is 1 unit. What is the sum of all distinct numbers x such that f(f(f(x)))=-3 ?

Guest Mar 10, 2019

#1**+1 **

First, we note that there are two points on the graph whose y-coordinates are -3. These are (-4,-3) and (0,-3). Therefore, if f(f(f(x)))=-3, then f(f(x)) equals -4 or 0. There are three points on the graph whose y-coordinates are -4 or 0. These are (-2,-4), (-6,0), and (2,0). Therefore, if f(f(x)) is -4 or 0, then f(x) equals -2, -6, or 2. There are four points on the graph whose y-coordinates are -2 or 2 (and none whose y-coordinate is -6). The x-coordinates of these points are not integers, but we can use the symmetry of the graph (with respect to the vertical line x=-2) to deduce that if these points are (x1,-2), (x1,-2), (x3,2), and (x4,2), then x1+x2=-4 and x3+x4=-4. Therefore, the sum of all four x-coordinates is -8.

Guest Mar 10, 2019

#2**+1 **

Mmm

A portion of the graph of y = f(x) is shown in red below, where f(x) is a quadratic function. The distance between grid lines is 1 unit. What is the sum of all distinct numbers x such that f(f(f(x)))=-3 ?

f(0) = -3 and f(-4)= -3

So now I need f(x)=0 and f(x)=-4

f(2)=0, f(-6)=0 f(-2)=-4

So now I need

f(x)=2, f(x)=-6 and f(x)=-2

f(2.8)=2, f(-6.8)=2 and none, and f(0.8)=-2 f(-4.8)=-2

f(2.8)=2

f(f(2.8)=f(2)=0

f(f(f(2.8))=f(f(2))=f(0)=-3

same for the others

so

the possible x values are 2.8, -6.8, 0.8 and -4.8 (all approximate)

sum = -8

Guest, the content of your answer may well be better than mine but it is so blocked together that it would be very difficult for people to decipher.

Presentation is important.

Melody Mar 10, 2019

#3**+1 **

f(f(f(x)) = - 3

Take the inverse of both sides

f^{-1} f(f(f(x)) = f^{-1} (-3) = - 4 and 0

f(f(x) = f^{-1} (-3) = - 4 and 0

Take the inverse of both sides again

f^{-1} f(f(x) ) = f^{-1}(-4) and f^{-1} f(f(x)) = f^{-1}(0)

f(x) = -2 and f(x) = -6 and f(x) = 2

Take the inverse one last time

f^{-1}f(x) = f^{-1}(-2) and f^{-1} f(x) = f^{-1}(-6) and f^{-1} f(x) = f^{-1}(2)

x = f^{-1}(-2) and x = f^{-1}(-6) and x = f^{-1} (2)

Since x = - 2 is a line of symmetry

Then f^{-1}(-2) can be written as -2 - a, -2 + a

And f^{-1}(2) = can be written as -2 - b, -2 + b

And f^{-1}(-6) = not on graph

So....the sum of the coordinates =

(-2 - a) + (-2 + a) + (-2 - b) + (-2 + b) = -8

CPhill Mar 10, 2019