Pre cal: is there a value of k such that the function has no zero?

Question

f(x)= 1/5 (x^5 - 20x^3 + 64x) + k

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Is there a value of k such that the function has no zero?

Answer 1

The answer is no because any fifth (or other odd) degree equation has to have at least one (real) zero.  Complex roots occur in pairs.

Answer 2

f(x) will have at least one zero no matter what the value of k.  Here is why:

For x negative with sufficiently large absolute value f(x)  will be negative, and for x sufficiently large and positive f(x) will be positive.  f is continuous, so by the intermediate value theorem there will be at least one value of x for which f(x) is zero, and this same argument applies no matter what the value of k.  The same applies to any odd degree polynomial, but not to even degree polynomials.