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Define

$$F(x)=\int_1^x \frac{1}{2\sqrt{t}-1} dt \quad \text{for all $x\ge 1$}.$$

Prove that if c>0, then there is a unique solution to the equation

$$F(x)=c, \quad x>1.$$

Attempt at a solution: I am not sure how to "prove" this. What I have so far is that $\int_1^x \frac{1}{2\sqrt{t}-1} dt= \sqrt{x}+\frac{1}{2} ln(2\sqrt{x}-1)-1$. If F(x)=c, and x>1, then F(x) will always be positive so clearly there will be a unique solution to F(x)=c...there must be a trick I am not seeing.

Marie
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1 Answers1

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Note that $$F'(x) = {1\over 2\sqrt{x}+1}$$ for $x > 1$.
The function $F$ is strictly increasing, and $F(0) = 0$. Note that as $x\to\infty$, $F(x) \uparrow\infty$.

Therefore, for all $c > 0$, $F(x) = c$ has a unique solution.

ncmathsadist
  • 49,383
  • Helped so much thank you! One quick question, how did you get the "+" on the bottom? I keep going through and am getting $\frac{1}{2\sqrt{x}-1}$ – Marie Oct 21 '13 at 00:52
  • You have $F'(x)$ behaving like ${1\over 2 \sqrt{x}$ at $\infty$, so $F$ is going to increase at $\infty$ like the square-root function. – ncmathsadist Oct 21 '13 at 01:19