The topic is combinations of functions.

Let $f=\frac{\sqrt{x}}{x}$ and $g = 2x-3$. Find the range of $f+g$.

$f+g = \frac{\sqrt{x}}{x} + 2x -3$.

Usually for complicated functions, you'd find the domain of the inverse, but this function does not pass the horizontal line test, therefore it does not have an inverse for me to find. The course I am helping out with requires the students to use calculators with no graphing capabilities.

EDIT: I know there's a minimum when $x=4^{-2/3}$ which results in $y \approx -0.619$ but I used calculus to solve. Since this is a precalculus course, I cannot teach that method.

Hope someone can provide assistance.

  • $\begingroup$ Are you allowed to use calculus? $\endgroup$
    – 3x89g2
    Sep 13, 2016 at 3:20
  • $\begingroup$ @Misakov: Unfortunately I am not. $\endgroup$ Sep 13, 2016 at 3:22
  • $\begingroup$ Range can't be that. At $x=1$, $f+g=0$ $\endgroup$ Sep 13, 2016 at 3:41
  • $\begingroup$ @WinVineeth You sure? Because $4^{-2/3}$ is approximately 0.4 EDITED $\endgroup$ Sep 13, 2016 at 3:43
  • 1
    $\begingroup$ Can you say something like- x rises twice as fast as sqrt(x). And hence, 1/sqrt(x) = 4x for minimum... something on these lines would be the best possible without using calculus. $\endgroup$ Sep 13, 2016 at 4:01

1 Answer 1


The minimum can be obtained by using AM-GM as follows: \begin{align*} \frac{1}{\sqrt{x}}+2x &= \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}+2x\\ &\geq 3\left(\frac{1}{2\sqrt{x}}\frac{1}{2\sqrt{x}}2x\right)^{1/3} \\ &= \frac{3}{2^{1/3}} \end{align*} The minimum is reached when $\frac{1}{2\sqrt{x}} = 2x$ or $x = 4^{-2/3}$

  • $\begingroup$ Note this only works because $\frac{1}{\sqrt x}$ is only defined for $x ≥ 0$. $\endgroup$
    – Toby Mak
    Aug 4, 2019 at 3:12

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