You need to identify the extreme points of the function, hence, you need to solve the equation `f'(x) = 0` , such that:

`f'(x) = (cx^4 - 2x^2 + 1)' => f'(x) = 4cx^3 - 4x`

`f'(x) = 0 <=> 4cx^3 - 4x = 0 => 4x(cx^2 - 1) =...

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You need to identify the extreme points of the function, hence, you need to solve the equation `f'(x) = 0` , such that:

`f'(x) = (cx^4 - 2x^2 + 1)' => f'(x) = 4cx^3 - 4x`

`f'(x) = 0 <=> 4cx^3 - 4x = 0 => 4x(cx^2 - 1) = 0 => {(4x = 0),(cx^2 - 1 = 0):} => {(x = 0),((cx - 1)(cx + 1) = 0):} => {(x = 0),(x = +-1/c):}`

Hence, if `c > 0` , th function reaches its minimum points at `x = +-1/c` . In the graph sketched below, c is considered equal to 1.

The function reaches its maximum points if `c < 0` , hence, the graph below illustrates the maximum point obtained for `c = -1` .