Posted by **Leanna** on Sunday, September 12, 2010 at 10:09pm.

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 -
**MathMate**, Sunday, September 12, 2010 at 11:30pm
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+3x^{6}-x^{9}=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 -
**MathMate**, Sunday, September 12, 2010 at 11:32pm
Typo : "rigorous" instead of vigorous.

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