I want to show that $\displaystyle 1 \leq \sqrt{1+x^3} \leq 1+x^3$. 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'm not sure what you mean. Are trying to say that $\displaystyle 1\leq \sqrt{1+x^3}$ is the same as $\displaystyle \sqrt{1+x^3}\leq 1+x^3$ ???
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, $\displaystyle x \geq 0$.
The excercise appears in a chapter on the fundamental theorem. One of the properties of integrals is $\displaystyle m(b-a)\leq \int_a^bf(x)dx \leq M(b-a)$
Where $\displaystyle m\leq f(x) \leq M$ and $\displaystyle x$ is in $\displaystyle [a,b]$.
This property is used in one of the proofs of the fundamental theorem.
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 $\displaystyle 1 + x^3 \le (1 + x^3)^2 $.
You know that if c is a positive number greater than or equal to 1, then $\displaystyle 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.