Find the intervals on which the following function is increasing and decreasing and as well as its concavity? f(x)=x^2(1+x^2)

I can derive the f' and f" but I'm having difficulties finding the values
f'(x)=2x+4x^3
f"(x)=2+12x^2

Do I just evaluate the f' and f" without equation the functions to 0?

1 Answer
Sharkbasket
Mar 25, 2018

#f(x)# is decreasing for #x<0#
#f(x)# is increasing for #x>0#
#f(x)# is concave upward for all #x#.

Explanation:

The question is asking you to find the intervals for which #f'(x)# and #f''(x)# are positive and negative.

As you probably already know:

If #f'(x)# > 0, #f(x)# is increasing at x.
If #f'(x)# = 0, #f(x)# has a horizontal tangent at x.
If #f'(x)# < 0, #f(x)# is decreasing at x.

If #f''(x)# > 0, #f(x)# is concave upward at x.
If #f''(x)# = 0, #f(x)# has an inflection point at x.
If #f''(x)# < 0, #f(x)# is concave downward at x.

#f(x)=x^2(1+x^2)=x^4+x^2#
#f'(x)=4x^3+2x#
#f''(x)=12x^2+2#

First let's evaluate #f'(x)#

#f'(x)=4x^3+2x=2x(2x^2+1)#

By examining the factored form of #f'(x)# we can see that #(2x^2+1)# is positive for all real values of x.

Therefore, #f'(x)# is only negative if #2x<0->x<0#.

#f(x)# is decreasing for #x<0#
#f(x)# is increasing for #x>0#

Now let's examine #f''(x)#.

#f''(x)=12x^2+2#

Once again by examination, we can see that #f''(x)# is positive for all real values of x. Therefore:

#f(x)# is concave upward for all #x#.

As visual proof, let's look at a graph of #f(x)#:

graph{x^4+x^2 [-2.574, 2.425, -0.33, 2.17]}

As expected, it is a parabola centred at #x=0#.

Related questions
Impact of this question
6056 views around the world
You can reuse this answer
Creative Commons License