What are the points of inflection, if any, of #f(x)= x^5 -2 x^3 - x^2-2 #?

1 Answer
Jul 30, 2018

The function has three Inflection points at
#x= -0.672 , x= -0.176 , x= 0.847 #.

Explanation:

#f(x) = x^5-2 x^3-x^2-2#

#f'(x) = 5 x^4- 6 x^2- 2 x#

# f''(x) = 20 x^3-12 x -2 ; f"(x)=0 #, critical points are

#x=-0.672, x= -0.176, x=0.847# Let’s select a convenient

number in the interval less than #-0.672#,

between #−0.672 and -0.176#, between #−0.176 and 0.847#

and greater than #0.847#. Let those be # −1, -0.5, 0.5 ,1#

respectively . #f^''(-1) ~~ (-) , f^''(-0.5) ~~ (+) # and

#f^''(0.5) ~~ (-) , f^''(1) ~~ (+) # , therefore , concave down

at #< -0.672# , concave up between #−0.672 and -0.176#,

concave down between # -0.176 and 0.847# and

concave up at #(-oo,-0.672)#. Hence the The function has three

Inflection points at #x= -0.672 , x= -0.176 , x= 0.847 #

graph{x^5-2x^3-x^2-2 [-20, 20, -10, 10]}

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