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Could anyone shine some light on this question please?

By considering $f'(x)$, show that $$f(x)=x^3 - 2$$ has exactly one root for $x$ greater than or equal to $0$.

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Hint: $f'(x)=3x^2\ge 0$ for all $x$.

Note that the function is increasing and $f(0)<0$ and $f(2)>0$, what can you conclude using continuity of $f$?

Side-note the roots of the equation are $2^{1/3}\cdot \omega ^i$ where $i=0,1,2$ and $\omega$ is the cube root of unity. i.e. there is only one real root.

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  • $\begingroup$ Thanks a lot for your help, appreciate it! $\endgroup$ – gary Oct 23 '14 at 20:23
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After differentiate, you have $f'> 0$ for $x>0$, so it's strictly increasing. Then by intermediate value theorem, there exists a root. By strictly increasing, the root is unique.

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  • $\begingroup$ Ah that's great thank you, it was the second part that you picked up on that I wasn't too sure about1 $\endgroup$ – gary Oct 23 '14 at 20:12
  • $\begingroup$ @gary $f(0)<0,f(3)>0$ and $f$ is continuous. The intermediate value theorem guarantees there is a root. $\endgroup$ – John Oct 23 '14 at 20:22
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By intermediate value theorem, you can see that $f(0) < 0$ and $f(2)>0$, therefore $f$ has a root $x_0$ between 0 and 2. Now suppose there exists another $x_1 \geq 0$ such that $f(x_1) = f(x_0) = 0$ By Rolle's Theorem, that would mean there exists $x_0\leq x_2 \leq x_1$ such that $f'(x_2) = 0$.

Now $f'(x) = 3x^2$. Note that $f'(x) = 0 \iff x = 0$, but $0$ isn't found in the interval we're considering.

Therefore, there can't be two roots of this polynomial function greater or equal than 0.

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Note that $f(0) < 0$. Take the derivative and note that it is positive on the entire interval. Conclude that f must cross the axis only once.

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