+130516 For the first one we have
f(x)=x^4-25x^2 ....factor and set to 0
x^4-25x^2 = 0
x^2 (x^2 - 25) = 0
x^2 (x + 5) (x- 5) = 0 and it's clear that the three roots are 0, -5 and 5
Here's the graph......https://www.desmos.com/calculator/g2ddxunylc
f(x)= x^3 - 14x^2 + 64x - 97
f ' (x) = 3x^2 - 28x + 64
Set to 0
3x^2 - 28x + 64 = 0 factor if possible
(3x -16)(x - 4) = 0
Setting each factor to 0 , we have that x = 16/3 and x = 4
To find if these produce mins or max's take the second derivative
f ''(x) = 6x - 28
At 16/3, we have .. f ''(16/3) = 6(16/3) - 28 = 32 - 28 = + so we have a min at x = 16/3
When x = 4, we have a max, because 6(4) - 28 = -4
Here's a graph of this one ........https://www.desmos.com/calculator/bxs8h9hfgb
f(x)= -x^3 + 4x^2 - 5
f ' (x) = -3x^2 + 8x factor and set to 0
x(3x - 8) = 0 and x = 0 and x = 8/3
Again, take the second derivative and plug in these critical points
f ''(x) =-6x + 8
f ''(0) = 8 = + which means a min at x = 0
f ''(-8/3) = -16 + 8 = -8 which means a max at x = 8/3
Here's a graph..........https://www.desmos.com/calculator/fpn5caadg5
