f(x)=x^4-2x^3?

Question

Find Relative extrema, Incresing and decreasing, absolute min and max, concave up and down, critical point

Answer 1

f(x) = x^4-2x^3

Use the first derivative to determine whether f'(x) is increasing or decreasing since f'(x) really represents the slope at every point on the line. i.e. If f'(x) is negative, the slope is negative, and the function would be decreasing.

Let's start by finding the derivative:

f'(x) = 4x^3-6x^2

The critical numbers (and also your local minima and maxima) occur when the slopes change from either negative to positive or positive to negative. Realize that this occurs when the slope equals zero.

Hence:

f'(x) = 0 = 4x^3-6x^2
0 = x^2(4x-6)
x = 0, x = 3/2

Since 0 and 3/2 are the critical values, we need to check if the slope is negative or positive for the intervals they give:

(−∞, 0), (0,3/2), and [3/2,∞)

We do this by choosing values besides the critical numbers that are contained within each interval:

(−∞, 0)
Choose x = -1:
f'(-1) = 4(-1)^3-6(-1)^2 = -10
f'(x) is negative, and f(x) is therefore decreasing

(0,3/2)
Choose x = 1:
f'(1) = 4(1)^3-6(1)^2 = -2
f'(x) is negative, and f(x) is therefore decreasing

(3/2,∞)
Choose x = 2:
f'(2) = 4(2)^3-6(2)^2 = 8
f'(x) is positive, and f(x) is therefore increasing

This shows that there is no local maximum and the local minimum occurs at x = 3/2.

In order to test for concavity, you need the second derivative. If f''(x) is negative, f(x) is concave down and if f''(x) is positive, f(x) is concave up.

Let's solve for the second derivative:


f'(x) = 4x^3-6x^2
f''(x) = 12x^2 -12x

Once more, we need to find the critical values:

f''(x) = 0 = 12x^2 -12x
f''(x) = 0 = 12x(x -1)
x = 0, x = 1

We again have three intervals (−∞, 0), (0,1), and (1,∞) with which to work.

(−∞, 0)
Choose x = -1
f''(-1) = 12(-1)^2 -12(-1) = 24
f''(x) is positive, and f(x) is therefore concave up

(0, 1)
Choose x = 1/2
f''(1/2) = 12(1/2)^2 -12(1/2) = -3
f''(x) is negative, and f(x) is therefore concave down

(1,∞)
Choose x = 2:
f''(2) = 12(2)^2 -12(2) = 24
f''(x) is positive, and f(x) is therefore concave up

Answer 2

f'(x)=4x^3-6x^2=2x^2(x-1)
f'(x)=0<-->x=0 or x=3/2
f'(x)>0 <----> XE(-Inf,0) and (3/2,Inf)
f is increasing in (-Inf,0] and [3/2,Inf)
f'(x)<0 <---> XE(0,3/2)
f is decreasing in[0,3/2]
Relat. max (0,0)
Rel. min (3/2,-27/16)
No abs min and max exist
since the values of f are (-Inf,Inf)
f''(x)=12x^2-12x=12x(x-1)
f''(x)>0<--> (-Inf,0) and ((1,Inf)
f concave up in (-Inf,0] and [1,Inf)
f''(x)<0<--> (0,1)
f conc.down in[0,1]
Critical points at x=0 and x=3/2

Answer 3

take derivative
4x^3-6x^2=0

x^2 (4x-6)=0

the roots are
x1=0
x2=3/2

Can you continue?


find the roots.

Answer 4

Is F part of this equation? Does it have a value? A little more info please!! I can get you the value of x no problem. Just explain what F is.