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
y=f(-1)
y=f(0)
y=f(1)
evaluate f(x) at those points to get your lines.
let f be the function defined by f(x)=3X^5 -5X^3 +2
a) on what interval is f increasing? b) on what interval is the graph of f concave upward?
c)Write the equation of each horizontal line tangent to the graph of f
4 answers
Oops. f'' = 30x(2x^2 - 1)
so -1/√2 < x < 0 or x > 1/√2
so -1/√2 < x < 0 or x > 1/√2
a) That would be where the derivative
f'(x) = 15x^4 -15x^2 > 0
x^2*(x^2-1) >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.
f'(x) = 15x^4 -15x^2 > 0
x^2*(x^2-1) >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.
So is the answer y=0, y=2, and y=4?
2 answers