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Math Help - Fundamental Theorem

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by adkinsjr View Post
    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
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    Quote Originally Posted by tonio View Post
    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 ???

    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.
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  4. #4
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    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
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    Quote Originally Posted by JG89 View Post
    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.
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  6. #6
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    Quote Originally Posted by adkinsjr View Post
    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 ???

    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
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