What are the absolute extrema of $f (x) = \frac{x - 1}{x^{2} + x + 2} \in [\infty, \infty]$ $f(x)=(x-1)/(x^2+x+2) in[oo,oo]$ ?

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What are the absolute extrema of $f (x) = \frac{x - 1}{x^{2} + x + 2} \in [\infty, \infty]$ $f(x)=(x-1)/(x^2+x+2) in[oo,oo]$ ?

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Answer 1 · 2016-10-11T15:55:56

At $x = - 1$ $x=-1$ the minimum
and at $x = 3$ $x=3$ the maximum.

Explanation:

$f (x) = \frac{x - 1}{x^{2} + x + 2}$ $f(x)=(x-1)/(x^2+x+2)$ has stationary points characterized by

$\frac{d f}{d x} = - \frac{(x - 3) (1 + x)}{{(2 + x + x^{2})}^{2}} = 0$ $(df)/(dx) = -((x-3) (1 + x))/(2 + x + x^2)^2=0$ so they are at

$x = - 1$ $x=-1$ and $x = 3$ $x=3$

Their characterization is made analyzing the signal of

$\frac{d^{2} f}{{d x}^{2}} = \frac{2 (x ((x - 3) x - 9)) - 1}{{(2 + x + x^{2})}^{3}}$ $(d^2f)/(dx^2) = (2 (x ((x-3) x-9))-1)/(2 + x + x^2)^3$ at those points.

$\frac{d^{2} f}{{d x}^{2}} (- 1) = 1 > 0 \to$ $(d^2f)/(dx^2)(-1)=1 > 0->$ relative minimum
$\frac{d^{2} f}{{d x}^{2}} (3) = - \frac{1}{49} < 0 \to$ $(d^2f)/(dx^2)(3)=-1/49 < 0->$ relative maximum.

Attached the function plot.

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What are the absolute extrema of $f (x) = \frac{x - 1}{x^{2} + x + 2} \in [\infty, \infty]$ $f(x)=(x-1)/(x^2+x+2) in[oo,oo]$ ?

1 Answer

Explanation:

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What are the absolute extrema of f(x)=x−1x2+x+2∈[∞,∞]f(x)=(x-1)/(x^2+x+2) in[oo,oo]?

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What are the absolute extrema of $f (x) = \frac{x - 1}{x^{2} + x + 2} \in [\infty, \infty]$ $f(x)=(x-1)/(x^2+x+2) in[oo,oo]$ ?