If f(x) = x(x + 2)^2, show that the point of inflection lies exactly halfway between the extremas.?

Answer 1

f(x)= x^3+4x^2+4x rewrite the original equation
f '(x)= 3x^2+8x+4 find its derivative
f '(x)=0 set it equal to zero
3x^2+8x+4=0 and solve
x= -2 and x=-2/3 answers= local extremas

to find the point of inflection set the second derivative equal to zero and solve
f ''(x)= 6x+8
set f ''(x)=0 set to zero and solve
x= -4/3 point of inflection

-4/3 is the half way point between -2 and - 2/3

Answer 2

We have f(x) = x^3 + 4x^2 + 4x,
f'(x) = 3x^2 + 8x + 4 and f"(x) = 6x + 8. Since f'(x) = 0 at x = -2/3 and x = -2, the extrema occur at
(-2/3, f(-2/3)) = (-2/3,-32/27) and (-2,f(-2)) = (-2,0). The midpoint of the extrema is
(-4/3.-16/27).
Since f"(x) = 0 at x = -4/3, the inflection is at
(-4/3,f(-4/3)) = (-4/3, -16/27).

Answer 3

f'(x) = 3x^2+8x+4
f'(x) = 0 when x = -2 and -(2/3)
from that we know the inflection point must be -4/3
f''(x) = 6x+8
f''(x) = 0 when x = -4/3