Explain why x - 3x^5 = 1/4, has at least two solutions within the interval [0,1]

Answer 1

x - 3x^5 = 1/4

-3x^5 +x - 1/4 = 0 has 2 change of signs By Descartes Method there's a maximum 2 real positive solutions

replace x by -x and

3x^5 -x -1/4 = 0 has 1 change of sign, so there is a maximum 1 negative real solution

that means 2 or 4 imaginary solutions, imaginary solutions always come in conjugate pairs

every 5th degree polynomial equation has 5 solutions, as many solutions as the highest degree term. although if the graph of f(x) is tangent to the x axis at a point, that zero has multiplity two, the solution repeats.

when x=0, f(x) = -3x^5 +x -1/4 = -1/4

when x =1 f(x) = -3+1-1/4 = -2 1/4

when x=1/2, f(x) = -3(1/2)^4 +1/2 -1/4 = -3/32 +1/4 = +5/32

there's two positive real solutions, one in the interval (0,5/32) and a second in the interval (5/32, 1)

f(2) is even more negative

f(0) = -1/4

f(-1) =3-1 -1/4=1 3/4

f(-2) = 48-2-1/4 = 45 3/4

somewhere between x=0 and -1 is a solution

where f(x) changes sign is a solution, which is where the graph of f(x) crosses the x axis

as x gets more negative f(x) is more positive,

so there is one negative solution

as x get more positive, larger than 1 f(x) gets even more negative.

so just 1 real solution, negative

but it has to be 1 or 3, as imaginary solutions come in pairs

there must be 0 or 2 positive real solutions

extrema are when f'(x) = 0 = 1-15x^4

x^4 = 1/15

x^2 = +/- sqr(1/15) = about +/-1/4

x = +/- 4th root of 1/15 = about +/- 1/2. one 4th root is a local max, the other a local min

if they're opposite signs a zero or solution is between them

a graphing calculator shows a local maximum at about x=1/2 and a local minimum about x=-1/2

the Desmos graphing calculator shows 3 real solutions, one negative, two positive, about x=1/4 and x=3/4 with a local maximum f(x) at about x=1/2, for f(x)=-3x^5+x-1/4

that seems to be the 3 real solutions.1 negative real, 2 positive real, and 2 imaginary solutions