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Let $$f(x)=\sqrt{9x^{2}+1}-3x-2 $$

  • Show that $f$ is bounded from below by $-2$ $$\forall x\in \mathbb{R}\quad f(x)>-2 $$
  • Is $-2$ the minimum value of $f$ ?

Indeed, let $x\in \mathbb{R}$ \begin{align} f(x)>-2 &\iff \sqrt{9x^{2}+1}-3x-2>-2\\ &\iff \sqrt{9x^{2}+1}>3x \\ &\implies 9x^{2}+1>9x^2 \\ &\implies 1>0 \end{align}

then $$\forall x\in \mathbb{R}\quad f(x)>-2 $$ which means $f$ is bounded from below by $-2$

  • Is $-2$ the minimum value of $f$ ?

I can't show that $-2$ is not a value of f(x) to say that $-2$ isn't minimum value of $f$

Beware: No differentiability

MathematicianByMistake
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Yacob
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    Note that any proposition, true or false, implies $1>0$, hence proving the implication $f(x)>-2 \implies 1>0$ does not help in the least. – Did Dec 16 '16 at 08:09
  • Did you plot the function? – Michael Hoppe Dec 16 '16 at 09:27
  • @MichaelHoppe yes http://www.wolframalpha.com/input/?i=plot+%7C+sqrt(9+x%5E2+%2B+1)+-+3+x+-+2 – Yacob Dec 16 '16 at 09:56
  • Are you confusing minimum and infimum? – Michael Hoppe Dec 16 '16 at 11:17
  • @MichaelHoppe yes i do would you tell me the difference – Yacob Dec 21 '16 at 15:22
  • @Yacob The infimum of a set $M$ is the greatest lower bound of $M$, if such a number exist. For example, the infimum of $A=(0,1)$ is $0$. Note here, that the infimum is not an element of $A$. Now if the infimum belongs to $M$ it is called the minimum of $M$ as in $[0,1)$. For your question: $-2$ is the infimum of $f$, but since there's no $x$ with $f(x)=-2$, we know $f$ has no minimum. – Michael Hoppe Dec 22 '16 at 08:18

3 Answers3

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Your $f(x)>-2$ argument is problematic because, as it is writtten, you can't go from $1>0$ in the bottom back to $f(x)>-2$. An easy fix is by replacing the last two $\Rightarrow$ by $\Leftarrow$'s.

Alternatively, you can also rephrase the argument as follows: because $9x^2+1>9x^2$, we have $$ f(x)>\sqrt{9x^2}-3x-2=3|x|-3x-2=3(|x|-x)-2\geq -2. $$ And because the leftmost inequality is strict, $2$ is not the minimum value of $f$.

yurnero
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Note that $$ \sqrt{9x^2+1}>3x $$ if and only if one of the two conditions : $$ x<0 \quad \mbox{or} \quad \begin{cases} x\ge 0\\ 9x^2+1>9x^2 \end{cases} $$ is true. And, since $9x^2+1>9x^2$ is true $\forall x \in \mathbb{R}$, we can conclude that the function is $f(x)> -2 \quad \forall x \in \mathbb{R}$.

The same calculations shows that the equation $f(x)=-2$ has no solutions, so $-2$ is not a minimum for the function.

Emilio Novati
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We have : $$f(x)=\sqrt{9x^{2}+1}-3x-2 $$ So $$ f'(x) = \frac{9x}{\sqrt{9x^2+1}} - 3$$ It is obvious that $f'(x)\lt0$ :

$f'(x) \lt 0 \iff \frac{9x}{\sqrt{9x^2+1}} \lt 3 \iff \frac{81x^2}{9x^2+1} \lt 9 \iff 0\lt9$ . Now we know that $f(x)$ is strictly decreasing . Now we calculate limit using L'Hôpital : $$ \lim_{x\to \infty}\sqrt{9x^{2}+1}-3x-2 = -2$$ Also we know $f(x) \neq -2$ as you calculate in your question Therefore we can conclude that $f(x)\gt -2$

S.H.W
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