Calculus Derivative Problem?
What is the Derivative of f(x) = 3x^4 + 4x^3 - 12x^2 + 10
a) Find all critical numbers of f(x)
b) Determine the open intervals(s) on which the function is increasing or decreasing.
c) Relative minimum(s)
d) Relative maximum(s)
Comments
a)
f(x) = 3x^4 + 4x^3 - 12x^2 + 10
f'(x)=12x^3+12x^2-24x
f'(x)=12x(x^2+x-2)=0
x=0
x^2+x-2=0
(x+2)(x-1)=0
x=1, x=-2
The critical numbers are x=0, x=1, x=-2
b)
Consider the intervals (-∞,-2),(-2,0),(0,1),(1,∞)
choose any one point from each of the intervals.
if f'(x) < 0 , f(x) is decreasing on that interval
if f'(x) > 0 , f(x) is increasing on that interval
f'(x)=12x^3+12x^2-24x
(-∞,-2): choose x=-3; f'(-3)= -144 < 0 : f(x) is decreasing on (-∞,-2)
(-2,0): choose x=-1; f'(-1)= 24 > 0 : f(x) is increasing on (-2,0)
(0,1): choose x=0.5; f'(0.5)= -7.5 < 0 : f(x) is decreasing on (0,1)
(1,∞): choose x=2; f'(2)=96 > 0 : f(x) is increasing on (1,∞)
c)
f''(x) =36x^2+24x-24
At x=0, f''(x) = -24 < 0, so f has a maximum at x=0
At x=-2, f''(x) = 72 > 0, so f has a minimum at x=-2
At x=1, f''(x) = 36 > 0 , so f has a minimum
Relative minimums at x=1,x=-2
f(1)=5
f(-2)=-22
(1,5),(-2,-22)
d)
Relative maximum at x=0
f(0) = 10
(0,10)
f'(x) = 12x^3 + 12x^2 - 24x
Find the critical values by setting f'(x) = 0 and solving.
12x^3 + 12x^2 - 24x = 0
12x(x^2 + x - 2) = 12x((x + 2)(x - 1) = 0
The critical values are x = -2, 0, and 1 (where relative minima or maxima occur).
Think of the end behavior of the 4th degree polynomial and how these critical values fit into what you know of the shape; the curve must have a relative minimum at -2, a relative maximum at 0, and a relative minimum at 1.
From this you can say the function is
decreasing on (-inf, -2)
increasing on (-2, 0)
decreasing on (0, 1)
increasing on (1, +inf)
A graphing calculator is a handy tool to check your answers, but you should not have to rely on it to answer problems like this.
f '(x)=12x^3 +12x^2 -24x
graph the derivative and use the calc/trace app to find relative minimums and maximums