Question

Find the critical points for each function. Use the derivative test to determine whether the critical point is a local maximum, local minimum, or neither.

y = x^4 - 8x^2

Answer 1

find the 1st and 2nd derivatives

f'(x) = 4x^3 - 16x

f"(x) = 12x^2 - 16

let f'(x) = 0 to find the stationary points

4x^3 - 16x = 0

4x(x^2 - 4) = 0

4x(x+2)(x-2)= 0

then startionary points at x = -2, 0, 2

Answer 2

Right, thats where I got up to but I don't know where to go from there

Answer 3

The answers are
local minima: (-2, -16), (2,-16)
local maximum: (0,0)

Answer 4

use the 2nd derivative test now

f'(c) = 0 and f"(c) > 0 then y = f(x) has a minimum turning point at x = c
f'(c) = 0 and f"(c) < 0 then y = f(x) has a maximum turning point at x = c
f'(c) = 0 and f"c) = 0 then y = f(x) has a horizontal point of inflexion at x = c

find f"(-2) = 48 - 16 >0 so a minimum
f"(0) = - 16 < 0 so a maximum
f"(2) = 48 - 16 > 0 so minimum

Answer 5

a point(s) of inflexion may exist at f"(x) = 0

so

12x^2 = 16

\[x = \pm 2/\sqrt{?}\]

Answer 6

?= 3