x^3-12x
my bad
Use the second derivative test to find the relative extrema of f(x) = x
3 − 12x.
3 answers
y = x^3 - 12x
y' = 3x^2 - 12
= 0 for either a max or min
3x^2 = 12
x^2 = 4
x = ±2
if x = 2, y = 8-24 = -16 ---> (2, -16)
if x = -2, y = -8 + 24 = +16 -->(-2,16)
now which is a max or min???
y'' = 6x
for x = 2, y'' = +12 , so at the point(2,-16) the curve is concave up
so (2,-16) is a minimum
for x = -2, y'' = -12, so at the point (-2,16) the curve is concave down
and (-2,16) is a maximum
https://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E3+-+12x
y' = 3x^2 - 12
= 0 for either a max or min
3x^2 = 12
x^2 = 4
x = ±2
if x = 2, y = 8-24 = -16 ---> (2, -16)
if x = -2, y = -8 + 24 = +16 -->(-2,16)
now which is a max or min???
y'' = 6x
for x = 2, y'' = +12 , so at the point(2,-16) the curve is concave up
so (2,-16) is a minimum
for x = -2, y'' = -12, so at the point (-2,16) the curve is concave down
and (-2,16) is a maximum
https://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E3+-+12x
max when y'=0 and y" > 0
min when y'=0 and y" < 0
y' = 3x^2 - 12 = 4(x^2-4)
y" = 6x-2 = 2(3x-1)
min when y'=0 and y" < 0
y' = 3x^2 - 12 = 4(x^2-4)
y" = 6x-2 = 2(3x-1)