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
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).
Typo : "rigorous" instead of vigorous.
To find the number of solutions to the equation f(x) = f^(-1)(x), we need to solve for x by equating the two functions.
Given that f(x) = 1 - x^3 and f^(-1) is the inverse of f, we can determine the inverse function by interchanging the roles of x and f(x) and solving for x.
Let y = f(x), so we have y = 1 - x^3.
To find the inverse, we interchange x and y and solve for x:
x = 1 - y^3.
Now, we set f(x) equal to its inverse, f^(-1)(x):
1 - x^3 = 1 - y^3.
From here, we can see that the equation is independent of x and y. We can solve for y^3 by subtracting 1 from both sides:
0 = y^3.
The only value of y that satisfies this equation is y = 0.
Substituting y = 0 back into the equation for the inverse function, we have:
x = 1 - (0)^3 = 1.
Therefore, the equation f(x) = f^(-1)(x) has one solution, x = 1. Thus, the answer is option b) one.