f(x) is equal to ax^2 - sqrt(2)
what is the value of a when f(f(sqrt(2)) = -sqrt(2)?
if f(x) is equal to ax^2 - sqrt(2)), then f(sqrt(2)) is equal to a * sqrt(2)^2 - sqrt(2).
you are simply replacing x with sqrt(2).
f(sqrt(2)) is equal to a * sqrt(2)^2 - sqrt(2) which is equal to a * 2 - sqrt(2) which is equal to 2a - sqrt(2).
you have f(sqrt(2)) is equal to 2a - sqrt(2).
f(f(sqrt(2))) is equal to f(2a - sqrt(2)).
you are simply replacing f(sqrt(2)) with 2a - sqrt(2)).
f(2a - sqrt(2)) is equal to a * (2a - sqrt(2))^2 - sqrt(2)
you are simply replacing x in f(x) = a * x^2 - sqrt(2) with (2a - sqrt(2)).
since (2a - sqrt(2))^2 is equal to 4a^2 - 4a*sqrt(2) + 2, your equation of:
f(2a - sqrt(2)) is equal to a * (2a - sqrt(2))^2 - sqrt(2)becomes:
f(2a - sqrt(2)) is equal to a * (4a^2 - 4a*sqrt(2) + 2) - sqrt(2).
you know two things.
1. a has to be positive.
2. the end result has to be - sqrt(2).
this end result can only occur if a * (4a^2 - 4a*sqrt(2) + 2) is equal to 0.
since a has to be positive, this can only occur if 4a^2 - 4a*sqrt(2) + 2 is equal to 0.
since this is a quadratic equation, set it equal to 0 and factor it using the quadratic formula.
you will get a = sqrt(2)/2.
here's the graph of your final equation of y = a * (4a^2 - 4a*sqrt(2) + 2) - sqrt(2).
you can see that the graph intersect with the graph of the line y = -sqrt(2) in two places.
one place is when a = 0 but this is invalid because a has to be positive.
the other place is when a is equal to sqrt(2)/2
sqrt(2)/2 is equal to .7071 on the graph.
-sqrt(2) is equal to -1.414
the solution is at the point (x,y) = .7071,-1.414).
