What are the points of inflection of $f(x)= (x^2 - 8)/(x+3)$ ?

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Answer 1 · 2017-08-11T07:01:32

There are no points of inflections

We need

$(u/v)'=(u'v-uv')/(v^2)$

Here,

$f(x)=(x^2-8)/(x+3)$

$u(x)=x^2-8$ , $=>$ , $u'(x)=2x$

$v(x)=x+3$ , $=>$ , $v'(x)=1$

Therefore, the first derivative is

$f'(x)=(2x*(x+3)-(x^2-8)*1)/(x+3)^2$

$=(2x^2+6x-x^2+8)/(x+3)^2$

$=(x^2+6x+8)/(x+3)^2$

And the second derivative is

$u(x)=x^2+6x+8$ , $=>$ , $u'(x)=2x+6$

$v(x)=(x+3)^2$ , $=>$ , $v'(x)=2(x+3)$

$f''(x)=((2x+6)(x+3)^2-2(x+3)(x^2+6x+8))/(x+3)^4$

$=((2x+6)(x+3)-2(x^2+6x+8))/(x+3)^3$

$=((2x^2+6x+6x+18)-(2x^2+12x+16))/(x+3)^3$

$=2/(x+3)^3$

The inflection points are when $f''(x)=0$

$2/(x+3)^3=0$ , $=>$ , $S=O/$

graph{(x^2-8)/(x+3) [-58.53, 58.55, -29.24, 29.23]}

What are the points of inflection of f(x)= (x^2 - 8)/(x+3) f(x)= (x^2 - 8)/(x+3) ?