# 009A Sample Final A, Problem 10

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10. Consider the function   ${\displaystyle f(x)=2x^{3}+4x+{\sqrt {2}}.}$
(a) Use the Intermediate Value Theorem to show that ${\displaystyle f(x)}$ has at least one zero.
(b) Use Rolle's Theorem to show that ${\displaystyle f(x)}$ has exactly one zero.

Foundations:
The Intermediate Value Theorem. If f is a continuous function on the interval [a,b], and if f(a) ≤ f(b), then for any y such that f(a) ≤ y ≤ f(b), then there exists a c ∈ [a,b] such that f(c) = y. Similarly, if f(a) ≥ f(b), then for any y such that f(a) ≥ y ≥ f(b), then there exists a c ∈ [a,b] such that f(c) = y.

In part (a) of this problem, as many others, we are trying to show that a root or zero exists. In order to apply the IVT, we need to note that the function is continuous, and then find an a and b such that, for example, f(a) < 0, while f(b) > 0.
Rolle's Theorem. If f is a continuous, real-valued function on the interval [a,b], and if f(a) = f(b), then there exists a c ∈ [a,b] such that f'(c) = 0.

For part (b), we will assume there are two roots, and show that it leads to a contradiction through Rolle's Theorem.

Solution:

Part (a):
We need to find two values a and b such that one is positive, and one is negative. Notice that f(0) = √2, which is greater than zero.
We can choose x = -1, to find f(-1) = -2 - 4 + √2, which is less than zero. Since f is clearly continuous, the IVT tells us there exists a c between -1 and 0 such that f(c) = 0.
Part (b):
Suppose there exists another root, say d. Then f(d) = f(c) = 0, so by Rolle's Theorem there exists some z ∈ [c,d] such that f'(z) = 0.
However, we know that f '(x) =6x2 + 4, which is greater than zero for all x. This contradicts that f'(z) = 0, so the assumption that another root exists must be incorrect.