Intermediate Value Theorem Question?

Question

The problem asks you to use the Intermediate Value Theorem to prove that a polynomial f(x)=1+2x+9x^2...3^101x^101 has a real zero.

How would you do this, especially with such a large number? Thanks for any help!

Answer 1

The previous answer was right but if it is very difficult to evaluate, just note that it is an odd polynomial with a positive leading term. Since it goes to infinity as x tends to infinity and goes to negative infinity if x tends to negative infinity and since polynomials are continuous then, you can be sure of at least one real zero (i.e. the graph should cross the x-axis), to prove this rigorously, you could transcribe this in terms of M and deltas.

Without calculus: we know that complex roots come in pairs, it is impossible to have all roots to be complex since by the fundamental theorem of algebra, it has 101 roots which is odd.

Answer 2

All you need to do is find one value of x where f(x) > 0 (e.g. x = 0, since the constant term is 1) and one value of x where f(x) < 0. Note that you don't need to evaluate f(x), you just need to show that it will be negative for some particular value of x. Then by the IVT there must be a zero in between the two points.