Is this graph right?

Question

Given f(x)=4x-x^3 determine the following: 

Domain: all real numbers 
y intercept: 0 
x intercept: 0, 2, -2 
1st derivative: 4-3x^2 
2nd derivative: -6x 
Critical Points +/- 2/sqrt(3) 
Interval f(x) increases: (-2/sqrt(3), 2sqrt(3)) 
Interval f(x) decreases: (-inf, -2/sqrt(3) U (2/sqrt(3), inf) 
Local maximum: 2/sqrt(3) 
Local minimum: -2/sqrt(3) 
Point of inflection: 0 
Interval concave up: (-inf, 0) 
Interval concave down: (0, inf)

Answer 1

The function f(x) = 4x - x³

f(x) = - x (x - 2) (x + 2)

1) Zeros of f(x) = - x (x - 2) (x + 2)

using zeros to graph the polynomial.

- x  = 0 , (x - 2) = 0 and (x + 2) = 0

x  = 0 , x  = -2 ,  x = 2.

Real zeros are x  intercepts of the graph.

2)Test points

Make the table of values to for the polnomial.

Here i test 3 points to determine whether the graph of polynomials lies above or below the x axis.

Choose values for x and find the corresponding values for y.

x	f(x) = 4x - x3	( x, y )
-1	y = 4(-1) - (-1)3= -3	(- 1, -3)
0	y = 4(0) - (0)3=  0	(0, 0)
1	y = 4(1) - (1)3 = 3	(1, 3)

3) End behavior f (x ) =  - x³ - 4x

Degree of polynomial is 3 and leading coefficient -1.

The graph of a polynomial function is always a smooth curve; that is, it has no breaks or corners.

All odd degree polynomials behave on thier ends like cubics.

All odd degree polynomials are either up on both ends and or down on both ends go of opposite directions..depending on whether the polynomial has, respectively, a positive or negative leading coefficient.

The above polynomial odd degree  polynomial with a negative leading coefficient .

So the graph start up and go down.

4)Graph

1.Draw a coordinate plane.

2.Plot the  intercepts coordinate points found in the table.

3.Then sketch the graph, connecting the points with a smooth curve.

The function f(x) = 4x - x³

Domain of the function is all real numbers.

y intercept is (0,0)

f(x) = 4x - x³

Apply derivative on each side with respect of x.

Apply the formula d/dx(xⁿ) = nx^n-1

First derivative f'(x) = 4 - 3x²

Apply derivative on each side with respect of x.

Second derivative is f''(x) = - 6x

To find critical points set derivative equals to 0.

4 - 3x²  = 0

3x²  = 4

x²  = 4/3

x = ± √(4/3)

Crtical points are x = ± 2/√3

Local maximum value is 2/√3 and local minimum value is -2/√3.

From the graph the function is increasing interval (-2/√3, 2/√3)

the function is decreasing interval (-∞,-2/√3) U (2/√3,∞)

Concave up (-∞,0) and concave down (0,∞)

To find inflection points set second derivative equals to 0.

-6x = 0

x = 0

Therefore, Point of inflection is zero.

From the graph we observe the given data is correct.