# Fundamental Theorem

• Nov 10th 2009, 11:09 AM
Fundamental Theorem
I want to show that $1 \leq \sqrt{1+x^3} \leq 1+x^3$. I'm not sure where to begin on this. It is an excercise in a chapter about the Fundamental Theorem of calculus. Any suggestions?
• Nov 10th 2009, 12:12 PM
tonio
Quote:

I want to show that $1 \leq \sqrt{1+x^3} \leq 1+x^3$. I'm not sure where to begin on this. It is an excercise in a chapter about the Fundamental Theorem of calculus. Any suggestions?

I suppose x is positive, and then the left inequality is trivial, and the right one is...exactly the same as the left one! Can you see it?

Tonio
• Nov 10th 2009, 12:58 PM
Quote:

Originally Posted by tonio
I suppose x is positive, and then the left inequality is trivial, and the right one is...exactly the same as the left one! Can you see it?

Tonio

I'm not sure what you mean. Are trying to say that $1\leq \sqrt{1+x^3}$ is the same as $\sqrt{1+x^3}\leq 1+x^3$ ???

(Thinking) I don't see that

I can see how the entire innequality is obviously true. I just don't see how to prove it formally. And yes, $x \geq 0$.

The excercise appears in a chapter on the fundamental theorem. One of the properties of integrals is $m(b-a)\leq \int_a^bf(x)dx \leq M(b-a)$

Where $m\leq f(x) \leq M$ and $x$ is in $[a,b]$.

This property is used in one of the proofs of the fundamental theorem.
• Nov 10th 2009, 02:59 PM
JG89
You don't need the FTC for this. The left hand inequality is obviously true for non-negative x. For the right hand inequality, square both sides. You get $1 + x^3 \le (1 + x^3)^2$.

You know that if c is a positive number greater than or equal to 1, then $c \le c^2$, right? So just put c = 1 + x^3
• Nov 10th 2009, 03:20 PM
Quote:

Originally Posted by JG89
You don't need the FTC for this. The left hand inequality is obviously true for non-negative x. For the right hand inequality, square both sides. You get $1 + x^3 \le (1 + x^3)^2$.

You know that if c is a positive number greater than or equal to 1, then $c \le c^2$, right? So just put c = 1 + x^3

That's what I was thinking. It's a very obvious inequality, I just thought it was strange that it was an excercise in a chapter on the fundamental theorem. I thought that perhaps someone might see a connetion between the two, because I don't. It's a very obvious algebraic statement.
• Nov 10th 2009, 07:08 PM
tonio
Quote:

I'm not sure what you mean. Are trying to say that $1\leq \sqrt{1+x^3}$ is the same as $\sqrt{1+x^3}\leq 1+x^3$ ???

(Thinking) I don't see that

I can see how the entire innequality is obviously true. I just don't see how to prove it formally. And yes, $x \geq 0$.

The excercise appears in a chapter on the fundamental theorem. One of the properties of integrals is $m(b-a)\leq \int_a^bf(x)dx \leq M(b-a)$

Where $m\leq f(x) \leq M$ and $x$ is in $[a,b]$.

This property is used in one of the proofs of the fundamental theorem.

For any poisitive number $a\,,\,\,\frac{a}{\sqrt{a}}=\sqrt{a}$ , so $\sqrt{1+x^3}\leq 1+x^3\Leftrightarrow 1\leq \frac{1+x^3}{\sqrt{1+x^3}}=\sqrt{1+x^3}$...the same left inequality we already had.

Tonio