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Show that $x^3+6x+3=0$ has exactly one solution in the interval [-3, 0].

Help me finish this solution.

Proof: (Existence) $f(x)=x^3+6x+3$, the function f is continuous on [-3,0], because it is a polynomial. Note that $f(-3)=-42<0$ and $f(0)=3>0$. By the Intermediate Value Theorem, there exist $cE(-3,0)$, such that $f(c)=0$.

Proof: (Uniqueness) Suppose there are two such solutions $a$ and $b$ where $f(a)=0$ and $f(b)=0$. Note that f is differentiable on (-3,0) because it is a polynomial. By Rolle's Value Theorem, there must exist a point $c$ between $a$ and $b$ such that $f'(c)=0$...

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    $\begingroup$ I'd think it's enough to note that the function is strictly increasing on the interval (and everywhere else for that matter) $\endgroup$
    – Dániel G.
    Commented Nov 3, 2014 at 20:59

2 Answers 2

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The derivative is always positive, and is therefore always increasing. If you are intent upon using Rolle's Theorem, then the statement that $f'(x) > 0 \implies f(x) \text{ is increasing}$ is proved using Rolle's Theorem.

Equivalently, if the derivative is always positive but there are two roots, then there is a point in the middle where the derivative is $0$, contradicting the fact that the derivative is always positive.

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Note that $f^{'}(t)=3x^{2}+6$ which is positive for all $x$. This implies the function is an unbounded increasing monotone - this alone guarantees that the root is unique.

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