Answer:
point (-1, 2)
Step-by-step explanation:
Your equation is shown here in point-slope form, so a point on the graph can be read from the given equation. (No further rewrite is necesary.)
The point-slope form is ...
y -k = m(x -h)
Comparing this to your given equation ...
y -2 = (-2/3)(x -(-1))
we see that ...
k = 2, m = -2/3, h = -1
The point on the graph is (h, k) = (-1, 2).
Y = (1 + x) / (1 + x^2)
y'
= [(1 + x^2)(1) - (1 + x)(2x)] / (1 + x^2)^2
= [1 + x^2 - 2x - 2x^2] / (1 + x^2)^2
= [-x^2 - 2x + 1] / (1 + x^2)^2
y''
= [(1 + x^2)^2 * (-2x - 2) - (-x^2 - 2x + 1)(2)(1 + x^2)(2x)] / (1 + x^2)^4
= [(1 + x^2)(-2x - 2) - (4x)(-x^2 - 2x + 1)] / (1 + x^2)^3
= [(-2x - 2x^3 - 2 - 2x^2) - (-4x^3 - 8x^2 + 4x)] / (1 + x^2)^3
= [-2x - 2x^3 - 2 - 2x^2 + 4x^3 + 8x^2 - 4x] / (1 + x^2)^3
= [2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3
Setting y'' to zero, we have:
y'' = 0
[2x^3 + 6x^2 - 6x - 2] / (1 + x^2)^3 = 0
(2x^3 + 6x^2 - 6x - 2) = 0
Using trial and error, you will realise that x = 1 is a root.
This means (x - 1) is a factor.
Dividing 2x^3 + 6x^2 - 6x - 2 by x - 1 using long division, you will have 2x^2 + 8x + 2.
2x^2 + 8x + 2
= 2(x^2 + 4x) + 2
= 2(x + 2)^2 - 2(2^2) + 2
= 2(x + 2)^2 - 8 + 2
= 2(x + 2)^2 - 6
Setting 2x^2 + 8x + 2 to zero, we have:
2(x + 2)^2 - 6 = 0
2(x + 2)^2 = 6
(x + 2)^2 = 3
x + 2 = sqrt(3) or = -sqrt(3)
x = -2 + sqrt(3) or x = -2 - sqrt(3)
Note that -2 - sqrt(3) < -2 + sqrt(3) < 1
We will choose random values belonging to each interval and test them out.
-5 < -2 - sqrt(3) < -2 < -2 + sqrt(3)
f''(-5) = [2(-5)^3 + 6(-5)^2 - 6(-5) - 2] / (1 + (-5)^2)^3 = -9/2197 < 0
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0
Note that one value is positive and the other is negative.
Thus, x = -2 - sqrt(3) is an inflection point.
-2 - sqrt(3) < -2 < -2 + sqrt(3) < 0 < 1
f''(-2) = [2(-2)^3 + 6(-2)^2 - 6(-2) - 2] / (1 + (-2)^2)^3 = 18/125 > 0
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0
Note that one value is positive and the other is negative.
Thus, x = -2 + sqrt(3) is also an inflection point.
-2 + sqrt(3) < 0 < 1 < 2
f''(0) = [2(0)^3 + 6(0)^2 - 6(0) - 2] / (1 + (0)^2)^3 = -2 < 0
f''(2) = [2(2)^3 + 6(2)^2 - 6(2) - 2] / (1 + (2)^2)^3 = 26/125 > 0
Note that one value is positive and the other is negative.
Thus, x = 1 is an inflection point.
Hence, we have three inflection points in total.
When x = -2 - sqrt(3), we have:
y
= (1 - 2 - sqrt(3)) / (1 + (-2 - sqrt(3))^2)
= (-1 - sqrt(3)) / (1 + 4 + 4sqrt(3) + 3)
= (-1 - sqrt(3)) / (8 + 4sqrt(3))
When x = -2 + sqrt(3), we have:
y
= (1 - 2 + sqrt(3)) / (1 + (-2 + sqrt(3))^2)
= (-1 + sqrt(3)) / (1 + 4 - 4sqrt(3) + 3)
= (-1 + sqrt(3)) / (8 - 4sqrt(3))
When x = 1, we have:
y
= (1 + 1) / (1 + 1^2)
= 2 / 2
= 1
Using the slope formula, we have:
(y - 1) / (x - 1) = [[(-1 + sqrt(3)) / (8 - 4sqrt(3))] - 1] / ( -2 + sqrt(3) - 1)
(y - 1) / (x - 1) = 1/4, which is the equation of the line which the inflection points at x = 1 and x = -2 + sqrt(3) lies on.
Note that I am skipping the intermediate steps for simplifying here, but the trick is to rationalise the denominator by multiplying a conjugate on both numerator and denominator.
Now, we just need to check that the inflection point at x = -2 - sqrt(3) lies on the same line as well.
L.H.S.
= [[(-1 - sqrt(3)) / (8 + 4sqrt(3))] - 1] / (-2 - sqrt(3) - 1)
= 1/4
= R.H.S.
Once again, I am skipping simplifying steps here.
<span>Anyway, this proves all three points of inflection lies on the same straight line.</span>
Answer:
112x m^2
Step-by-step explanation:
The area of a trapezoid is given by
A = 1/2 ( b1+ b2) h
where b1 and b2 are the bases
A = 1/2 ( 11x+17x) * 8
= 1/2 ( 28x)8
= 112x m^2
Answer:
$135
Step-by-step explanation:
180 decrease 25% =
180 × (1 - 25%) = 180 × (1 - 0.25) = 135
