REALLY hard math question worth alot of marks..dont get it ..help please?

Question

PLEASE HELP ME!!! I NEED TO KNOW THE ANSWER FOR AN EXAM REVIEW..PLEASE HELP!!!!


f(x)= 2x³-x²+6x-3

1)determine x and y intercepts;
2)Determine co-orinates of critical points
3)Determine co-orinates of points of inflection
4)State intervals on increase/decrease
5)STate concavity

Answer 1

f(x) = 2x^3 - x^2 + 6x - 3

1) Determine the x and y intercepts.

By definition, the y-intercepts are found when x = 0, so we let x = 0.

f(0) = 2(0)^3 - 0^2 + 6(0) - 3
f(0) = -3
So the y-intercept is -3.

To find the x-intercepts, let f(x) = 0. Then
0 = 2x^3 - x^2 + 6x - 3

Fortunately for us, this can be factored by grouping.

0 = x^2(2x - 1) + 3(2x - 1)
0 = (x^2 + 3)(2x - 1)

Now, we equate each of these to 0. x^2 + 3 will give no solution, but 2x - 1 = 0 yields the solution x = 1/2. Therefore, our x-intercept is x = 1/2.

2) To find the coordinates of the critical points, we need to first solve for the first derivative, and then make it 0.

f'(x) = 6x^2 - 2x + 6

Setting f'(x) to 0,

0 = 6x^2 - 2x + 6.

Dividing by 2 will yield
0 = 3x^2 - x + 3.

To solve this now, we use the quadratic formula. But upon doing this, we will learn that there are no solutions. That means we have no critical points whatsoever.

3) To find the coordinates of the points of inflection, we take the second derivative. Note that the first derivative is

f'(x) = 6x^2 - 2x + 6. Therefore,
f''(x) = 12x - 2.

Setting f''(x) to 0, we get

0 = 12x - 2
-12x = -2, so
x = 1/6

To find the coordinates of this potential point of inflection, the coordinates will be (1/6, f(1/6)). Solving for f(1/6), we get

f(1/6) = 2(1/6)^3 - (1/6)^2 + 6(1/6) - 3
f(1/6) = 2(1/216) - (1/36) + 1 - 3
f(1/6) = 1/108 - 1/36 - 2
f(1/6) = 1/[12*9] - 1/[12*3] - 2
f(1/6) = 1/[12*9] - 3/[12*9] - 2[12*9]/[12*9]
f(1/6) = 1/108 - 3/108 - 216/108
f(1/6) = -218/108 = -109/54

Therefore, the coordinates of the point of inflection is
(1/6, -109/54)

4) To find the intervals of increase/decrease, refer back to the first derivative.
f'(x) = 6x^2 - 2x + 6

Since we've obtained no critical values, this function is either going to be always increasing or always decreasing. All we have to do is simply test ANY value for x in f'(x), and determine whether it is positive or negative. Test x = 0. Then
f'(0) = 0 - 0 + 6 = 6, which is positive.
Therefore, f(x) is ALWAYS increasing, i.e.
f(x) is increasing from (-infinity, infinity)

5) To determine concavity, we refer to the second derivative again.

f''(x) = 12x - 2

Recall that our critical point for the second derivative is 1/6. That means we test at value LESS than it, and determine if it is positive or negative, and then test a value GREATER than it, to determine if it's positive or negative. Positive will imply concave up; negative will imply concave down.

Test 0: then, f''(0) = 12(0) - 2 = -2, which is negative.
Therefore, f(x) is concave down on (-infinity, 1/6).

Test 1: then f''(1) = 12(1) - 2 = 10, which is positive.
Therefore, f(x) is concave up on (1/6, infinity)

Answer 2

1:
Find the x intercepts. Set f(x) = 0
Take a look here on factoring cubic equations:
http://library.thinkquest.org/C0110248/algebra/cubiceqn.htm
0 = (2x-1)*(x² +3)
So, x intercepts is at 0.5 (and sqrt(-3) but you are most likeley not expected to mess with complex numbers)

Now, find the Y intercepts. Set x = 0
f(0) = 0 - 0 + 0 -3
f(0) = -3
-3 is your Y intercept

2) Find the critical points. You have to find the first derivative of the function:
f ' (x) = 6x² - 2x + 6
Now set f ' (x) equal to 0 (since at critical points the slope is zero ... and the first derivative of the function is the equation to find the slope of the function at a specified point)
0 = 2 * ( 3x² - x - 3) in order for it to be 0, then 3x² - x - 3 = 0
3x² - x - 3 = 0
You will notice that is not possible, so there are NO critical points.

3) find f ''(x)
= 12x-2
set it equal to 0
0 = 12x-2
2 = 12x
x = 1/6
Now test to check if it really is a point of inflection:
f ''(0) = -2
f ''(1) = 10
That means the concavity changed at those points; we conclude there is a point of inflection at (1/6, f(1/6)) = (1/6, -109/54)

4) Since we know there are no critical points, the function will always either be increasing or decreasing.
Find the slope of the function at any point, say 0
f ' (0) = 6*(0)² - 2*(0) + 6
f ' (0) = 6
Since the slope is positive, and we learned it is ALWAYS positive,
f increases from (-∞, ∞).

Answer 3

1) set x = 0, f(x=0) = - 3
f(x) = 0 =2*x^3 -x^2 + 6*x -3

2x*(x^2 + 3) + -1(x^2+3) = (2x-1)(x^2+3)
y = 0 when x is either {1/2, sqrt(3)i, -sqrt(3)i}

2) critical pts where f'(x) = 0 or does not exist

f'(x) = 6x^2 -2x + 6 = 0
divide by 6
x^2 - (1/3)x + 1 = 0

x^2 -(1/3)x + 1/9 - 1/9 + 1 = 0
(x-1/3)^2 = -8/9 , this leads to complex number answers, therefore the function has no pts where f'(x) = 0 where x is a real number.

3) f''(x) = 12x - 2, inflection pt at x = 1/6, concave up for X>6, concave down for X<1/6

it increases for all x, with no critical pts.

You can get free math/stat software with R, use the polyroot function and the plot function.

You enter as series of values using x<- seq(-4,4,by=0.1)