How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $f (x) = \frac{x^{2} + 1}{x^{2} - 4}$ $f(x)=(x^2+1)/(x^2-4)$ ?

Question

How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $f (x) = \frac{x^{2} + 1}{x^{2} - 4}$ $f(x)=(x^2+1)/(x^2-4)$ ?

1 Answer

Impact of this question

1678 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-18T10:09:46

There are no points of inflection.
The discontinuities are when $x = - 2$ $x=-2$ and $x = 2$ $x=2$
$f (x)$ $f(x)$ is concave up for $x \in] - \infty, - 2 [\cup] 2 . + \infty [$ $x in ]-oo,-2[ uu]2.+oo[$
$f (x)$ $f(x)$ is concave down for $x \in] - 2, 2 [$ $x in ]-2,2[$

Explanation:

As we cannot divide by $0$ $0$ , $x \neq - 2$ $x!=-2$ and $x \neq 2$ $x!=2$

The domain of $f (x)$ $f(x)$ is $d_f(x)=RR-{-2,2}$

Therefore, the discontinuities are when $x=-2$ and $x=2$

The derivative of a quotient is

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

We calculate the first derivative of $f(x)$

$u=x^2+1$ , $=>$ , $u'=2x$

$v=x^2-4$ , $=>$ , $v'=2x$

$f'(x)=(2x(x^2-4)-2x(x^2+1))/(x^2-4)^2$

$=(2x(x^2-4-x^2-1))/(x^2-4)^2$

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

We can calculate the second derivative of $f(x)$

$u=-10x$ , $=>$ , $u'=-10$

$v=(x^2-4)^2$ , $=>$ , $v'=2(x^2-4)*2x=4x(x^2-4)$

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

$=((x^2-4)(-10x^2+40+40x^2))/(x^2-4)^4$

$=(30x^2+40)/(x^2-4)^3$

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

The points of inflexion are when, $f''(x)=0$

$f''(x)!=0$ , there are no points of inflection.

Let build the chart for concavities

$color(white)(aaaa)$ $Intervals$ $color(white)(aaaa)$ $]-oo,-2[$ $color(white)(aaaa)$ $]-2,2[$ $color(white)(aaaa)$ $]2,+oo[$

$color(white)(aaaa)$ $x+2$ $color(white)(aaaaaaaaaaa)$ $-$ $color(white)(aaaaaaaaaaa)$ $+$ $color(white)(aaaaaaa)$ $+$

$color(white)(aaaa)$ $x-2$ $color(white)(aaaaaaaaaaa)$ $-$ $color(white)(aaaaaaaaaaa)$ $-$ $color(white)(aaaaaaa)$ $+$

$color(white)(aaaa)$ $Sign f''(x)$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaaaaaaaaa)$ $-$ $color(white)(aaaaaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaaaaaa)$ $uu$ $color(white)(aaaaaaaaaaa)$ $nn$ $color(white)(aaaaaaaa)$ $uu$

Therefore,

$f(x)$ is concave up for $x in ]-oo,-2[ uu]2.+oo[$

$f(x)$ is concave down for $x in ]-2,2[$

Science

Math

Humanities

... and beyond

How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $f (x) = \frac{x^{2} + 1}{x^{2} - 4}$ $f(x)=(x^2+1)/(x^2-4)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for f(x)=(x^2+1)/(x^2-4)f(x)=x2+1x2−4f(x)=(x^2+1)/(x^2-4)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for $f (x) = \frac{x^{2} + 1}{x^{2} - 4}$ $f(x)=(x^2+1)/(x^2-4)$ ?