relative extrema...?

Question

Use the second derivative test to identify the relative extrema of f(x) = (x^2 - 6x + 8)^2

Answer 1

f(x) = (x^2-6x+8)^2

Find the first derivative to obtain the critical points
f'(x) = 2(x^2-6x+8) (2x-6)
f'(x) = 2(x^2-6x+8) 2(x-3)
** f'(x) = 4(x^2-6x+8) (x-3)

** is what we will take second derivative of..
but first find the critical points by finding when f'(x) is 0

f'(x) = 0
4(x^2-6x+8) (x-3) = 0

Factor some more
4(x-4)(x-2)(x-3) = 0

crits at x= 2,3,4

Take second derivative so we can determine whether these are max or mins

f'(x) = 4(x^2-6x+8) (x-3)
f''(x) = 4 [ (x^2-6x+8) *1 + (x-3)(2x-6)]
f''(x) = 4 [ x^2-6x+8 + 2x^2 -12x + 18]
f''(x) = 12x^2-72x+104

Evaluate points
f''(2) = 8
f''(3) = -4
f''(4) = 8

Second derivative test says
f''(x) > 0 = min
f"(x) < 0 = max

So max at x=3
mins at x =2 and x =4

Answer 2

extrema of f(x) = (x^2 - 6x + 8)^2

df/dx = 2(x^2 - 6x + 8)(2x - 6)
df/dx = 0
(2x - 6) = 0 and x = 3
x^2 - 6x + 8 = 0 = (x - 4)(x - 2) = 0 and x = 2, x = 4
There are three points of interest: x =2,3 and 4

d^2f/dx^2 = 2[(x^2 - 6x + 8)(2) + (2x - 6)(2x - 6)]
d^2f/dx^2 = 2[2x^2 - 12x + 16 + 4x^2 - 24x + 36]
d^2f/dx^2 = 2[6x^2 - 36x + 52] = 4[3x^2 - 18x + 26]

at x = 2: d^2f/dx^2 = 8 > 0 so a minimum
at x = 3: d^2f/dx^2 = -4 < 0 so a maximum
at x = 4: d^2f/dx^2 = 8 > 0 so a minimum

Note: evaluating the second derivatives in detail.
4[3x^2 - 18x + 26] at x = 2
4[3*2^2 - 18(2) + 26] 4[12 - 36 + 26] = 4[2] = 8

this can also be written as 12x^2 - 72x + 104 but I prefer factoring out the 4 since to me this makes for easier calcs and less prone to errors.

Answer 3

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