# Let $x$ be a fixed real number. Prove: if $x^5 + 2x^3 + x < 50$, then $x < 2$

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.

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

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?

• So that would be a proof by example ? Jul 2 at 12:40
• @David What do you mean by "proof by example"? Jul 2 at 12:41
• 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 $x^5 + 2 * x ^ 3 + x > 50$ Jul 2 at 12:44

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.

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 $$f(x) < c \implies x.
Here, $$f(2) = 50$$. So $$f(x) < 50 \implies x<2$$

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.