I know f' has to = zero.
Is the answer 2?
(x=0,−1)
How many critical values does the function
f(x) = 3x^4 + 4x^3
have?
2 answers
3x^4 + 4x^3 =0
x^3(3x+4) = 0
x = 0 or x = -4/3
looks like two x-intercepts
f ' (x) = 12x^3 + 12x^2
= 0 for a max/min of f(x)
12x^3 + 12x^2 = 0
x^2(12x + 12) = 0
x = 0 , f(0) = 0
x = -1, f(-1) = 3 + 4(-1) = -1
looks like (-1, -1) is a minimum , and we will consider (0,0) later
f ''(x) = 36x^2 + 24x
= 0 at a point of inflection
x(36x + 24) = 0
x = 0 or x = -24/36 = -2/3
so (0,0) and (-2/3, ....) are points of inflection
confirmation:
http://www.wolframalpha.com/input/?i=plot+y+%3D+3x%5E4+%2B+4x%5E3
x^3(3x+4) = 0
x = 0 or x = -4/3
looks like two x-intercepts
f ' (x) = 12x^3 + 12x^2
= 0 for a max/min of f(x)
12x^3 + 12x^2 = 0
x^2(12x + 12) = 0
x = 0 , f(0) = 0
x = -1, f(-1) = 3 + 4(-1) = -1
looks like (-1, -1) is a minimum , and we will consider (0,0) later
f ''(x) = 36x^2 + 24x
= 0 at a point of inflection
x(36x + 24) = 0
x = 0 or x = -24/36 = -2/3
so (0,0) and (-2/3, ....) are points of inflection
confirmation:
http://www.wolframalpha.com/input/?i=plot+y+%3D+3x%5E4+%2B+4x%5E3
5 answers