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We have this inequality :

$1-x^2-\log x \ge 0$

How we can solve this inequality ? (Note : It is natural $\log$)

(In general , I have problem with inequalities involving logarithms and exponents functions.)

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  • $\begingroup$ what kind of $\log$ it is? $\endgroup$ – Dr. Sonnhard Graubner Nov 25 '16 at 11:56
  • $\begingroup$ It is natural $\log$ . $\endgroup$ – S.H.W Nov 25 '16 at 11:56
  • $\begingroup$ Consider $f(x)=1-x^2-\log(x)$.Compute $f'(x)$ and analyze it. $\endgroup$ – Claude Leibovici Nov 25 '16 at 11:58
  • $\begingroup$ @ClaudeLeibovici Are you sure derivatives are necesary here? I think it's easier to just look at what happens to the expression on $(0,1)$ and what happens on $(1,\infty)$. $\endgroup$ – 5xum Nov 25 '16 at 11:59
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    $\begingroup$ @5xum. Yes for sure. However, the derivative is quite interestin to look at. Cheers :-) $\endgroup$ – Claude Leibovici Nov 25 '16 at 12:01
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Let $x > 1$. Then $1-x^2<0$ and $\log x>0$. In this case the inequality is false.

Let $0<x < 1$. Then $1-x^2 > 0$ and and $\log x < 0$. In this case the inequality is true.

The case $x=1$ should be clear.

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  • $\begingroup$ I know it is obvious that intersection point is $x = 1$ but is there any algebraic way to find out this point ? In the other words , I want to solve $1-x^2 - \log x = 0$ $\endgroup$ – S.H.W Nov 25 '16 at 12:26
  • $\begingroup$ Let $f(x):=1-x^2-\log x$. Then $f(1)=0$. Now suppose that there is $x_0$ such that $x_0>0$, $x_0 \ne 1$ and $f(x_0)=0$. By Rolle, there is $x_1$ such that $f'(x_1)=0$. This leeds to $2x_1^2+1=0$ Contadiction $\endgroup$ – Fred Nov 25 '16 at 12:31
  • $\begingroup$ Thank you a lot , but I have another question . What is your mean by this ($:=$) symbol ? $\endgroup$ – S.H.W Nov 25 '16 at 12:48
  • $\begingroup$ $f(x):=1-x^2-\log x$ means: the function $f$ is defined by .... $\endgroup$ – Fred Nov 25 '16 at 14:43
  • $\begingroup$ So , What is the difference between $:=$ and $=$ ? $\endgroup$ – S.H.W Nov 25 '16 at 16:58
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If the $\log$ denotes the natural logarithm (i.e., base $e$), then

Hint:

  • What is the sign of $\log x$?
  • When $\log x$ is negative, what is the sign of the expression?
  • What about when $\log x$ is positive?
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Notice that we must have $x > 0$ for $\log x$ to be defined. Next, notice that $1 - x^2 - \log x = 0$ if and only if $1 - x^2 \geq \log x$.

The function $f_1(x) = 1 - x^2$ is strictly decreasing for the interval $(0;\infty)$ while $f_2(x) = \log x$ is strictly increasing for the same interval. We have that $\lim_{x \rightarrow 0} f_1(x) = 1$, while $\lim_{x \rightarrow 0} f_2(x) = -\infty$.

The graphs of $f_1$ and $f_2$ intersect at $x=1$, since $1 - 1^2 = \log 1 = 0$. We must therefore have that $0 < x \leq 1$ is the required solution.

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  • $\begingroup$ I know it is obvious that intersection point is $x = 1$ but is there any algebraic way to find out this point ? In the other words , I want to solve $1-x^2 - \log x = 0$ $\endgroup$ – S.H.W Nov 25 '16 at 12:25
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if we have $$f(x)=1-x^2-ln(x)$$ then we get $$\lim_{ x \to 0}f(x)=+\infty$$ and $$f'(x)$$ is given by $$f'(x)=-2x-\frac{1}{x}$$ for $$x>0$$ and $$\lim_{x \to \infty}f(x)=-\infty$$ thus we have only one intersection point with the $$x$$ axis whis is given by $x=1$

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  • $\begingroup$ I know it is obvious that intersection point is $x = 1$ but is there any algebraic way to find out this point ? $\endgroup$ – S.H.W Nov 25 '16 at 12:20
  • $\begingroup$ In the other words , I want to solve $1-x^2 - \log x = 0$ $\endgroup$ – S.H.W Nov 25 '16 at 12:22

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