Each X and Y in the equations don't have a number in front of them so they are all considered 1.
D = 1*(-1) - 1*(-1) = -1 -1 = -2
The answer is -2
The roots of the polynomial <span><span>x^3 </span>− 2<span>x^2 </span>− 4x + 2</span> are:
<span><span>x1 </span>= 0.42801</span>
<span><span>x2 </span>= −1.51414</span>
<span><span>x3 </span>= 3.08613</span>
x1 and x2 are in the desired interval [-2, 2]
f'(x) = 3x^2 - 4x - 4
so we have:
3x^2 - 4x - 4 = 0
<span>x = ( 4 +- </span><span>√(16 + 48) </span>)/6
x_1 = -4/6 = -0.66
x_ 2 = 2
According to Rolle's theorem, we have one point in between:
x1 = 0.42801 and x2 = −1.51414
where f'(x) = 0, and that is <span>x_1 = -0.66</span>
so we see that Rolle's theorem holds in our function.
<h3>Answer: y = (3/2)x + 0</h3>
This is the same as y = (3/2)x
=============================================================
Work Shown:
Find the slope of the line through (x1,y1) = (-2,-3) and (x2,y2) = (2,3)
m = (y2 - y1)/(x2 - x1)
m = (3 - (-3))/(2 - (-2))
m = (3 + 3)/(2 + 2)
m = 6/4
m = 3/2
The slope is the fraction 3/2. This is going to be in front of the x, or to the left of the x.
----------
Plug m = 3/2 and (x1,y1) = (-2,-3) into the point slope formula. Solve for y.
y - y1 = m(x - x1)
y - (-3) = (3/2)(x - (-2))
y + 3 = (3/2)(x + 2)
y + 3 = (3/2)x + (3/2)(2)
y + 3 = (3/2)*x + 3
y + 3 - 3 = (3/2)*x + 3 - 3
y = (3/2)x + 0
The y intercept is zero. This matches up with the fact the graph crosses the y axis at y = 0.
This problem mean you have to find the point that the Domain and Range is suit (I'm sorry English is my second language)
Example:
1) G ; 2) Y
If you want me to do the rest of it tell me Okay ^^
