polynomial functions and inequalities

Question

a) A polynomial function that has no real zeros will always be exclusively positive or negative (i.e. it will always be either all above or all below the x-axis – eg. f(x) = x2 + 2 has no real roots and it is always positive).

If a polynomial has real zeros, then if it is positive over any given interval, does it necessarily follow that it will be negative over the next sequential interval (i.e. if it is positive between the first and second root, does it necessarily become negative between the second and third root)? Explain using proper mathematical terminology.

b) provide a graph that illustrates your answer

Answer 1

to have real zeroes or roots, there must be an x intercept, where the curve intersects the x axis

if y is always positive or always negative, it never crosses the x axis, and the only roots are imaginary

y = x^2 + any positive number will always be above the x axis

y =-x^2 minus any positive number will always be below the x axis

or view it with the quadratic formula

for y = ax^2 +bx + c,

x = -b/2a + or - (1/2a)sqr(b^2 -4ac)

of the discriminate <0, the roots are both imaginary

if the discriminate > 0, the root(s) are real

for a linear equation, y=ax +b, there is no real solution if a=0 and b does not equal zero. There are "surreal" solutions, plus or negative infinity

for polynomials of degree 3 or more, try Descartes' rule, on the number of sign changes in the terms of the polynomial

or take a derivative, set it = 0 and solve for the maximum or minimum point. IF the maximum is < 0, there are not real solutions. IF the minimum is >0, there are no real solutions.

for example, y = 2x^2 + 2 has derivative y' = 2x = 0, x = 0 (0,2) is the minimum point, above the x axis, so there are no real solutions

It's an upward opening parabola if the coefficient of the x term is positive, downward opening if the coefficient is negative

for a 3rd, 5th, or odd degree polynomial, there is always at least one real solution, as it crosses the x axis, since there is no max or min. y approaches infinity as x approaches infinity, and y approaches -infinity as x apporaches -infinity, if the coefficient of the leading term is positive.

even degreed polynomials all have either an absolute minimum or absolute maximum. If the absolute minimum is > 0, there are only imaginary roots. If the absolute maximum is <0, there are only imaginary solutions.

there are as many roots as the degree of the polynomial. any imaginary roots come in conjugate pairs. so for an odd numbered degree polynomial there will be at least one real root, or an odd number of real roots. For even degreed polynomials, there may be zero real roots. or an even number of real roots

there are as many roots as the degree of the polynomial, (if you count a root with multiplicity 2 or more as being 2 or more roots). For example, a 101 degree polynomial has 101 roots, at least one real, or any odd number of real roots, with the rest imaginary.

y = x^101 has one real root, x = 0 with multiplicity 101. So you can say it has 101 roots, all identical.

y = x^2 has one real root, x=0 with multiplicity 2

a polynomial of degree 7 has potentially 7-1 = 6 turning points, which are each relative, local minimums or local maximums, all may be above or below the x axis, leaving only 1 real root, but potentially 7 real roots if each turning point involves crossing the x axis.

consider y = x^2 + 2. the discriminant is b^2 -4ac = 0 - 4(1)(2) = -8 < 0. Since it's less than zero, there are no real roots. Just a conjugate pair of imaginary roots. + or - (1/2)sqr(-8) = + or - isqr2 = about +1.14i

It's an upward opening parabola with minimum point or vertex (0,2) never intersecting the x axis

y=(x+2)(x)(x-2)= x^3 -4x has 3 zeros: -2, 0 and +2

It's concave down from - infinity to 0 and cocave up from 0 to +infinity. 0 is both a zero and an inflection point, where the concavity changes. y is positive for -2 < x < 0, and y is negative from 0 < x < +2

for y = -x^3 +4x there the same 3 zeros, but y is negative between x = -2 and x =0, and y is positive between x =0 and x= +2. It's concave up in the 1st interval and concave down in the 2nd interval.

It's concave up when the 2nd derivative is positive, concave down when the 2nd derivative is negative, and an inflection point (the dividing point where the concavity changes) when the 2nd derivative = 0

if f"(x) = 0, then that x value is an inflection point, but it isn't always an x intercept, root or zero.

f(x) = 2 is a flat line, where f'(0) =0 and f"(0) = 0, there is no x intercept, and no up or down concavity.

draw a "W" shaped or UU shapped curve, symmetric about the y axis, with 3 zeros, -2, 0 and +2, but the 0 zero has multiplicity of 2. At x=0, the curves is tangent to the x axis. It's a 4 degree polynomial: y = x^4 - 4x^2. the concavity is up on the left side, up on the right side, but down in the middle. But the y value is always non-positive, from -2 to 0 and from 0 to -2, never changing from negative to positive, except to the right of x=+2 and to the left of x=-2. So that's a counterexample to the apparent claim that each successive interval will change y from one sign to the other. It's not necessarily true that the y value will change signs over successive intervals of zeros.