Is $f(x)=(x^2+2x-2)/(2x-4)$ increasing or decreasing at $x=0$ ?

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Answer 1 · 2015-12-14T12:26:51

The function is decreasing.

If $f'(0)<0$ , then $f(x)$ is decreasing when $x=0$ .
If $f'(0)>0$ , then $f(x)$ is increasing when $x=0$ .

Find $f'(x)$ using the quotient rule.

$f'(x)=((2x-4)d/dx[x^2+2x-2]-(x^2+2x-2)d/dx[2x-4])/((2x-4)^2$

$f'(x)=((2x-4)(2x+2)-2(x^2+2x-2))/(4(x-2)^2)$

$f'(x)=(4x^2+4x-8x-8-2x^2-4x+4)/(4(x-2)^2$

$f'(x)=(2x^2-8x-4)/(4(x-2)^2$

$f'(x)=(x^2-4x-2)/(2(x-2)^2$

$f'(0)=(-2)/8=-1/4$

Since $-1/4<0$ , $f(x)$ is decreasing when $x=0$ .

Is f(x)=(x^2+2x-2)/(2x-4)f(x)=(x^2+2x-2)/(2x-4) increasing or decreasing at x=0x=0?