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$$y=x+\frac{1}{x-4}$$ I tried to find range of this function as below $$y(x-4)=x(x-4)+1\\yx-4y=x^2-4x+1\\x^2-x(4+y)+(4y+1)=0\\\Delta \geq 0 \\(4+y)^2-4(4y+1)\geq 0 \\(y-4)^2-4\geq 0\\|y-4|\geq 2\\y\geq 6 \cup y\leq2$$ this usuall way. but I am interested in finall answer... is the other idea to find function range ?

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use this fact : $|a+\frac 1a|\geq 2$ $$\quad{y=x+\frac{1}{x-4}\\y-4=x+\frac{1}{x-4}-4\\y-4=(x-4)+\frac{1}{x-4}=a+\frac 1a\\|y-4|\geq 2}$$

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Clearly for $|x|$ large, $y$ is large and has the same sign as $x$. For $x=4+h$ where $h$ is small, $y$ has the same sign as $h$ and is as large as we like in absolute value by making $h$ small. We can find the turning points by $$y'=1-\frac 1{(x-4)^2}=0, y''=\frac 2{(x-4)^3}$$ i.e. at $x-4=\pm 1$ with attention to the sign of $y''$.

When $x=3$ we have a local maximum $y=2$. For $x=5$ we have a local minimum $y=6$. These extreme points belong to their respective branches, and we therefore hit no values with $2\lt y\lt 6$ since one branch has $y \le 2$ and the other $y\ge 6$. Strictly we also need the simple observation that $y$ does take all other positive and negative real values (e.g. use the intermediate value theorem, or solve explicitly).

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Let $z:= x-4,$ and $z \ne 0$.

$y = z + 1/z +4$.

1) $z \gt 0:$

AM -GM:

$z+1/z \ge 2.$

Hence: $y \in [6,\infty)$.

2) $ z\lt 0$.

Above reasoning for $-z$ (positive),i.e.

$-z + 1/(-z) \ge 2$,

leads to:

$y \in (-\infty, 2]$.

Altogether :

Range $(y) = (-\infty, 2]\cup[4,\infty)$.

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