Posted by Yoona on Sunday, November 20, 2011 at 2:33pm.
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.
Calculous - let f be the function defined by f(x)=12x^2/3 -4X a)find the ...
CALCULUs - For all x in the interval [-11,13] the function f is defined by x^3(x...
Math - For x [14,13] the function f is defined by f(x)=(x^3)(x+6)^4 On ...
Honors Pre-Calculus - The prompt says "Graph each function. Determine the ...
Algebra - Im having a hard time figuring out this problem: Use the function, y...
AP Calculous - let f be the function defined by |x-1|+2 for X<1 f(x)= ax^...
increasing decreasing calc - f(x)=x^3=12x-24 f'(x)=3x^2+12=0 3x^2=-12 x^2=-4...
Calculus - Answer the following questions for the function f(x) = sin^2(x/3) ...
Calculus - f(x) = sin^2(x/2) defined on the interval [ -5.683185, 1.270796]. ...
Math - A function f is defined on the interval [0,4], and its derivative is e^...
For Further Reading