Help me out, please. Thank you!!
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You are correct. The answer is choice D
The only way for g(x) to be differentiable at x = 0 is for two things to happen
(1) g(x) is continuous at x = 0
(2) g ' (x) is continuous at x = 0
To satisfy property (1) above, the value of b must be 1. This can be found by plugging x = 0 into each piece of the piecewise function and solving for b.
So the piecewise function becomes
after plugging in b = 1
--------------------------------
Now differentiate each piece with respect to x to get
The first piece of g ' (x) is always going to be equal to 1. The second piece is equal to zero when x = 0
Because -sin(x) = -sin(0) = 0
So there's this disconnect on g ' (x) meaning its not continuous
Therefore, the value b = 1 will not work.
So there are no values of b that work to satisfy property (1) and property (2) mentioned at the top.
The only way for g(x) to be differentiable at x = 0 is for two things to happen
(1) g(x) is continuous at x = 0
(2) g ' (x) is continuous at x = 0
To satisfy property (1) above, the value of b must be 1. This can be found by plugging x = 0 into each piece of the piecewise function and solving for b.
So the piecewise function becomes
after plugging in b = 1
--------------------------------
Now differentiate each piece with respect to x to get
The first piece of g ' (x) is always going to be equal to 1. The second piece is equal to zero when x = 0
Because -sin(x) = -sin(0) = 0
So there's this disconnect on g ' (x) meaning its not continuous
Therefore, the value b = 1 will not work.
So there are no values of b that work to satisfy property (1) and property (2) mentioned at the top.
A parabolic function's key characteristic is either having 2 x-intercepts or 2 y-intercepts. That is the reason why the standard form of parabolic functions are:
(x-h)^2 = +/- 4a(y-k) or (y-k)^2 = +/- 4a(x-h), where
(h,k) is the coordinates of the vertex
4a is the lactus rectum
a is the distance from the focus to the vertex
This is also called vertex form because the vertex (h,k) is grouped according to their variable.
Since we don't know any of those parameters, we'll just have to graph the data points given as shown in the picture. From this data alone, we can see that the parabola has two x-intercepts, x=-4 and x=-2. Since it has 2 roots, the parabola is a quadratic equation. Its equation should be
y = (x+4)(x+2)
Expanding the right side
y = x²+4x+2x+8
y = x²+6x+8
Rearrange the equation such that all x terms are on one side of the equation
x²+6x+___=y-8+___
The blank is designated for the missing terms to complete the square. Through completing the squares method, you can express the left side of the equation into (x-h)² form. This is done by taking the middle term, dividing it by two, and squaring it. So, (6/2)²=9. Therefore, you put 9 to the 2 blanks. The equation is unchanged because you add 9 to both sides of the equation.
The final equation is
x²+6x+9=y-8+9
(x+3)²=y+1
(x-h)^2 = +/- 4a(y-k) or (y-k)^2 = +/- 4a(x-h), where
(h,k) is the coordinates of the vertex
4a is the lactus rectum
a is the distance from the focus to the vertex
This is also called vertex form because the vertex (h,k) is grouped according to their variable.
Since we don't know any of those parameters, we'll just have to graph the data points given as shown in the picture. From this data alone, we can see that the parabola has two x-intercepts, x=-4 and x=-2. Since it has 2 roots, the parabola is a quadratic equation. Its equation should be
y = (x+4)(x+2)
Expanding the right side
y = x²+4x+2x+8
y = x²+6x+8
Rearrange the equation such that all x terms are on one side of the equation
x²+6x+___=y-8+___
The blank is designated for the missing terms to complete the square. Through completing the squares method, you can express the left side of the equation into (x-h)² form. This is done by taking the middle term, dividing it by two, and squaring it. So, (6/2)²=9. Therefore, you put 9 to the 2 blanks. The equation is unchanged because you add 9 to both sides of the equation.
The final equation is
x²+6x+9=y-8+9
(x+3)²=y+1