Asked by Jacob
The function f(x)=x 3 +1 4 x−1 4 is a monotonically increasing function, hence it is injective (one-to-one), so its inverse function exists and is well defined. How many points of intersection are there, between the function f(x) and its inverse f −1 (x) ?
Answers
Answered by
Reiny
I assume you meant:
f(x) = x^3 + 14x - 14
(I don't understand why you would leave spaces between the digits)
so y = x^3 +14x - 14
its inverse is x = y^3 + 14y - 14
to intersect them, solve the two equations.
so
y = (y^3 + 14y - 14)^3 + 14(y^3 + 14y - 14) - 14
that looks like a nasty mess to solve, BUT
I noticed in the original, if x = 1 , I get y = 1
and in the inverse if y = 1, I get x = 1
so we lucked out to find the intersection point as (1,1)
Wolfram confirmed this.
http://www.wolframalpha.com/input/?i=solve+y+%3D+x%5E3+%2B+14x+-+14+%3B+x+%3D+y%5E3+%2B+14y+-+14
f(x) = x^3 + 14x - 14
(I don't understand why you would leave spaces between the digits)
so y = x^3 +14x - 14
its inverse is x = y^3 + 14y - 14
to intersect them, solve the two equations.
so
y = (y^3 + 14y - 14)^3 + 14(y^3 + 14y - 14) - 14
that looks like a nasty mess to solve, BUT
I noticed in the original, if x = 1 , I get y = 1
and in the inverse if y = 1, I get x = 1
so we lucked out to find the intersection point as (1,1)
Wolfram confirmed this.
http://www.wolframalpha.com/input/?i=solve+y+%3D+x%5E3+%2B+14x+-+14+%3B+x+%3D+y%5E3+%2B+14y+-+14
Answered by
Sahil
it is not 14 it is 1/4
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