Is $f(x)=(x^2+2x-6)/(2x+1)$ increasing or decreasing at $x=0$ ?

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Answer 1 · 2017-01-19T03:30:01

$uarr$ . Look for x = o, on the graph that is a hyperbola, with asynptotes $y =x/2+3/4 and 2x+1=0$ .

By actual division,

$f=x/2+3/4-(27/4)/(2x+1)$ and

$f'=1/2+(27/2)/(2x+1)^2=1/2+27/8=31/8>0$ , at x = 0.

So, f $uarr$ , at $(0, -6)$ .

See the graph.

Note that the graph is a hyperbola, with asynptotes

$y =x/2+3/4 and 2x+1=0$ . These meet at the

center $(-1/2, 1/2)$

graph{(y(2x+1)-x^2-2x+6)(x^2+(y+6)^2-.1)=0 [-20, 20, -10, 10]}

Is f(x)=(x^2+2x-6)/(2x+1)f(x)=(x^2+2x-6)/(2x+1) increasing or decreasing at x=0x=0?