Determine whether the function f(x)=xx2+2 is concave up or concave down and its intervals?

Step 1We have that f(x)=xx2+2 calculating its second derivative we find thatd2f(x)d2x=2x(x2−6)(x2+2)3So we need to see how the signs change of 2x(x2−6) as x goes from −∞ to +∞So from (−∞, −6] we have that f″(x)&lt;0from [−6, 0] we have that f″(x)&gt;0from [0, 6] we have that f″(x)&lt;0from [6, +∞) we have that f″(x)&gt;0In order to determine concavity we use the following theoremConcavity Theorem:If the function f is twice differentiable at x=c, then the graph of f is concave upward at (c; f(c)) if f″(c)&gt;0 and concave downward if f″(c)&lt;0

Expert answers to "Determine whether the function f(x)=xx2+2 is concave up..."

Reuben Brennan

Answered question

2022-03-26

Determine whether the function

f (x) = \frac{x}{x^{2} + 2}

is concave up or concave down and its intervals?

Answer & Explanation

clarkchica44klt

Beginner2022-03-27Added 17 answers

Step 1
We have that

f (x) = \frac{x}{x^{2} + 2}

calculating its second derivative we find that

d^{2} \frac{f (x)}{d^{2} x} = \frac{2 x (x^{2} - 6)}{{(x^{2} + 2)}^{3}}

So we need to see how the signs change of

2 x (x^{2} - 6)

as x goes from

- \infty

+ \infty

So from

(- \infty, - \sqrt{6}]

we have that

f^{″} (x) < 0

from

[- \sqrt{6}, 0]

we have that

f^{″} (x) > 0

from

[0, \sqrt{6}]

we have that

f^{″} (x) < 0

from

[\sqrt{6}, + \infty)

we have that

f^{″} (x) > 0

In order to determine concavity we use the following theorem
Concavity Theorem:
If the function f is twice differentiable at

x = c

, then the graph of f is concave upward at

(c; f (c))

f^{″} (c) > 0

and concave downward if

f^{″} (c) < 0

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?