Calculus challenge.?

Question

Let f(x) be a polynomial with real coefficients which has a real root. Show that there is some real number p so that f(p) = f '(p).

Hi Steve - It might be possible that there is an inductive proof. Certainly if h(x) = a x^n for n >= 2, then h(0)=h'(0), but ultimately this has to be shown for all polynomials which satisfy the requirements, i.e. showing that g(x) has the right properties.

Hi Steve - possibly, but I suspect that continued iteration might not be helpful in this case. The function g(x) doesn't necessarily satisfy the same properties as f(x), e.g. if f(x) = x^3 + 1, then g(x) = x^2 - x + 1, which has no real zeroes.

A more direct proof does exist which doesn't rely on induction.

Answer 1

Let g(x) = f(x) - f'(x). Then the problem can be restated as saying that g(x) has at least one real root. If f(x) is a polynomial of degree n, where n≠0, then f'(x) is a polynomial of degree n-1, and so the difference of the two polynomials will be a polynomial of degree n. We consider two cases:

Suppose n is odd. Then g(x) has odd degree, and all polynomials of odd degree have a real root, so we are done.

Suppose n is even. Then f(x) has either an absolute minimum or an absolute maximum. Suppose without loss of generality that it has an absolute maximum at c (the arguments for polynomials with an absolute minimum are essentially the same). Then at that point f'(c) = 0, so g(c) = f(c). Since f(x) has a real root, f(c) ≥ 0. If f(c) = 0 then g(c) = 0 and we are done. Otherwise, consider that the x^n term of f(x) dominates the terms of f'(x), so [x→∞]lim g(x) = -∞. Thus, g(x) is negative at at least one point, and since g(x) is everywhere continuous, then by the intermediate value theorem it must have at least one zero. Q.E.D.

Note that in the special case where n=0, f(x) is a constant function, and thus does not have a root, so it does not meet the conditions of the theorem. Finally, in the case where f(x) = 0 (which is usually considered to be a polynomial of indeterminate degree, rather than degree zero, primarily to avoid exceptions to the fundamental theorem of algebra), f(x) is everywhere equal to f'(x).

Answer 2

Let f(x) be any polynomial in R. Then, if f has real root z, f(x) = (x-z)g(x) for some polynomial g(x). so f'(x) = g(x)+(x-z)g'(x). Setting these equal, we have (x-z)g(x) - g(x) - (x-z)g'(x) = 0. What conditions on g(x) would allow the above equation to have a real root? Is it generally true if h(x) = ax^n, then h(0) = h'(0) = 0? Could induction prove that if g(x) is of the form h(x) above, the theorem holds?

Steve

Answer 3

2) lim x->3 [(x-one million)^2*(x-3)]/[(x+one million)^2*(x-3)] = lim x->3 [(x-one million)^2]/[(x+one million)^2] = [(3-one million)^2]/[(3+one million)^2] =4/sixteen=one million/4 3) lim x->2 [?(3x-2)-?(x+2)]/(x-2) = lim x->2 [?(3x-2)-?(x+2)]/(x-2) * [?(3x-2)+?(x+2)]/[?(3x-2)+?(x+2)] = lim x->2 [(3x-2)-(x+2)]/(x-2)[?(3x-2)+?(x+2)] = lim x->2 [2(x-2)]/(x-2)[?(3x-2)+?(x+2)] = lim x->2 2/[?(3x-2)+?(x+2)] = 2/[?(6-2)+?(2+2)] = 2/4=one million/2 4) lim q->0 [one million-cos2q]/[4q^2] = lim q->0 [one million-(one million-sin^2 q)]/[4q^2] = lim q->0 (sin^2 q)/4q^2 = one million/4 6) lim x->? (x-(one million/x))^2 = lim x->? x^2 - 2 - (one million/x)^2 = one million sorry it extremely is all i will do