# Calculus..Need Help Soon

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If f(x)=1-x^3 and f^-1 is the inverse of f, how many solutions does the equation f(x)=f^-1(x) have?

a)none b)one c)three d)five e)six

• Calculus..Need Help Soon - ,

Work out the inverse of f(x), which should be:
f-1(x) = (1-x)(1/3)
Equate
f(x)=f-1(x) to get
(1-x³)=(1-x)(1/3)
Take cube on both sides:
(1-x³)³ = (1-x)
Expand and cancel "1" on each side to give
x-3x³+3x6-x9=0 ...(1)

This is where it gets a little fussy.
Examine the signs of the coefficients equation and apply des Cartes rule of signs to get a minimum of three positive roots corresponding the three changes of sign.

In addition, we know that zero is a root which is non-positive. So the total number of roots (by the des Cartes rule) is 3+2k, where k is at least 1 (because zero is a root), meaning that the number of roots is either 5, 7 or 9. Of these, only 5 is a given choice.

I hope someone else could give a more vigorous solution.

(The solution to equation one actually yields 5 real and 4 complex zeroes).

• Calculus.-corr - ,

Typo : "rigorous" instead of vigorous.