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This is the problem: Let $x$ be a fixed real number. Prove: if $x^5 + 2x^3 + x < 50$, then $x < 2$. It's from the book "An introduction to mathematical proofs" by Nicholas A. Loehr.

I tried to prove it using a direct proof and then trying to prove that $x < 2$ by algebra. This is my work:

Proof (Direct Proof). Assume $x^5 + 2x^3 + x < 50$. I must prove $x < 2$.

By algebra,

$x^5 + 2x^3+x < 50$

$x(x^4 + 2x^2+1) < 50$

I can tell that it has something to do with the quadratic formula, but I don't really know where to start.

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  • $\begingroup$ Are you familiar with proof by contradiction? $\endgroup$
    – fwd
    Jul 2 at 12:35
  • $\begingroup$ @fwd yes. I'm familiar with direct proof, proof by contrapositive, proof of AND and IFF-statements $\endgroup$
    – David
    Jul 2 at 12:36
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    $\begingroup$ wouldn't that be contra-positive? $\endgroup$ Jul 2 at 12:37
  • $\begingroup$ You can consider two cases: when $x < 0$ and when $0 \le x <2$. $\endgroup$ Jul 2 at 12:42

4 Answers 4

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Note that $2^5+2\times2^3+2=50$, so for all $x>2$ you have $x^5+2\times x^3+x=50$. Can you conclude?

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  • $\begingroup$ So that would be a proof by example ? $\endgroup$
    – David
    Jul 2 at 12:40
  • $\begingroup$ @David What do you mean by "proof by example"? $\endgroup$ Jul 2 at 12:41
  • $\begingroup$ There are multiple types of proofs. I was trying to prove this using a direct proof. Someone in the comments mentioned that I could use a proof by contrapositive to solve this. In your proof you've given a value to $x$ and showed that $x^5 + 2*x^3 + x = 50$ therefore, for all $x > 2$ you will have $x^5 + 2 * x ^ 3 + x > 50$ $\endgroup$
    – David
    Jul 2 at 12:44
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You can find the zeroes of $x^5 + 2x^3 + x$ using your factorization. For $x^4 + 2x^2 + 1$, you can solve $u^2 + 2u + 1 = 0$, where $u = x^2$. Once you solve for all the zeroes, you know that a polynomial can only be positive or negative between the zeroes. That is the Intermediate Value Theorem.

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Let $f(x) = x^5 + 2x^3 + x$
So, $\frac{df}{dx} = 5x^4 + 6x^2 + 1$
Clearly, $\frac{df}{dx} > 0 \;\; \forall x \in \mathbb{R}$. So the function is strictly increasing.
This means if $f(x_0) = c$ for some $x_0 \in \mathbb{R}$ then $f(x) < c \implies x<x_0$.
Here, $f(2) = 50$. So $f(x) < 50 \implies x<2$

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Expanding @VioletFlame’ answer.
Since $x^5,x^3$ and $x$ are all increasing functions, we have $f(x)=x^5+2x^3+x$ as an increasing function in $\mathbb R$. Also note that $f(2)=50$.

Thus, if $x_1>2$ for some $x_1$, then $f(x_1)>f(2)=50$, which is against pur hypothesis that $f(x)<50$. Thus our assumption that $x>2$ has given rise to a contradiction. Thus our assumption must be false. Hence proved by way of contradiction.

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