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I was wondering if this was a valid existence proof for the following:

"there exists a unique real number solution to the equation $x^3 + x^2 - 1 = 0$ between $x = 2/3$ and $x = 1$"

Proof: Assume to the contrary that there are two real number solutions to the equation $x^3 + x^2 - 1 = 0$ between $x \in [2/3,1]$.

Let $a,b$ be the solution to $x^3 + x^2 - 1 = 0$. We may further assume that $2/3 < a < b < 1$ since $f(2/3) = -7/27$ and $f(1) = 1$ and so $f(x)$ is continuous on the interval $[2/3,1]$. Since $a^3 + a^2 - 1 < b^3 + b^2 - 1$ but since $a,b$ is a solution, then $a^3 + a^2 - 1 = 0$ and $b^3 + b^2 - 1 = 0$ which is a contradiction.

Does this seem like a valid proof ? The solution to this practice problem uses the intermediate value theorem but such a method was also given as an example in the textbook along with the latter approach.

I would greatly appreciate it if someone could verify.

Thank you.

PutsandCalls
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  • You probably want to say "let $a,b$ be distinct solutions to $x^2+x-1=0$" in the fourth paragraph. You might want to also include the reasons regarding why you think $a^3 + a^2 - 1 < b^3 + b^2 - 1$ is true as well. – Bernard W Jun 24 '16 at 07:25
  • @BernardWojcik You're right, that is a good idea. Thank you – PutsandCalls Jun 24 '16 at 07:29
  • Show that at the endpoints of the interval one endpoint is less than zero and one endpoint is greater than zero. If this is true, then there is at least one root in the interval (intermediate value theorem). If there are two zeros in the interval, there must be a stationary point in the interval. Rolle's theorem. There are no stationary points inside the interval. – Doug M Jun 24 '16 at 07:30
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    You say, $\frac 23<a<b<1 \implies a^3 + a^2 - 1 < b^3 + b^2 - 1$ but you don't say why. – Doug M Jun 24 '16 at 07:36
  • @Doug M I made a change but would saying that $f(x)$ is continuous on $[-2/3,1]$ given that it is a polynomial function and that $f(-2/3) = -7/27$ and $f(1) = 1$. Since 0 is between $f(-2/3)$ and $f(1)$, then $f(x)$ is increasing be a valid reason ? – PutsandCalls Jun 24 '16 at 07:45

3 Answers3

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Take $f(x)=x^3+x^2-1$. As you have already verified $f(2/3).f(1)<0$, it only indicates that $f(x)$ has atleast one root in the interval $(2/3,1)$. Now, to reject the possibility of more than one root (obviously odd), you just need that the derivative $f'(x)\neq 0$ in $(2/3,1)$.Hope, it helps.

Nitin Uniyal
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You have shown that when there are two real number solutions, there is a contradiction.

Have you shown that at least a solution exists? Intermediate value theorem is helpful here.

Siong Thye Goh
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Just prove $f(\frac {2}{3})f(1)<0$ which will automatically proves that there exist a root of $f(x)$ in $(\frac{2}{3},1)$ because $f(x)$ is contineous and differentiable in $R$.

Vineet Mangal
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