f is increasing when f' is positive
f' = 15x^4 - 15x^2 = 15x^2 (x^2-1)
So, f' > 0 when |x| > 1
f is concave upward when f'' is positive
f'' = 60x^3 - 30x = 30x(2x-1)
So, f'' > 0 when x < 0 or x > 1/2
Horizontal lines have slope=0. So, we want places where f'(x) = 0
15x^2 (x^2 - 1) = 0
x = -1, 0, 1
The horizontal lines are
evaluate f(x) at those points to get your lines.
Oops. f'' = 30x(2x^2 - 1)
so -1/√2 < x < 0 or x > 1/√2
a) That would be where the derivative
f'(x) = 15x^4 -15x^2 > 0
Since x^2 must be positive or zero,
(x+1)(x-1) > 0
x > 1 or x<-1
b) That would be where f"(x) > 0
c) Horizontal tangents would be where f'(x) = 0.
Find those x and y values.
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