What are the critical values, if any, of $f(x)=(x^2-5x)/(x^2-2)$ ?

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Answer 1 · 2015-11-04T18:07:56

$f$ has no critical values.

$f(x)=(x^2-5x)/(x^2-2)$

Note that the domain of $f$ is all real numbers except $+-sqrt2$

$f'(x) = ((2x-5)(x^2-2)-(x^2-5x)(2x))/(x^2-2)^2$

$= (2x^3-5x^2-4x+10-2x^3+10x^2)/(x^2-2)^2$

$= (5x^2-4x+10)/(x^2-2)^2$

$f'(x)$ is not defined for $x=+-sqrt2$ , but those numbers are not in the domain of $f$ , so they are not critical values for $f$ .

$f'(x)=0$ when $5x^2-4x+10 =0$ . The discriminant of this quadratic is negative, so there are no real zeros.

$f$ has no critical numbers.

What are the critical values, if any, of f(x)=(x^2-5x)/(x^2-2)f(x)=(x^2-5x)/(x^2-2)?