4x
d/dx 2x2+3x+7=4x+3
d/dx(1/2x) = -1/(2x2)
Since this question is in the calculus section, I'm assuming you know how to take the derivative. We know that y = -2x2 + 2x + 3 is a parabola, so it has one vertex, which is a minimum. We can use the first derivative test to find this extreme point.First, take the derivative:y' = -4x + 2Next, set y' equal to zero:0 = -4x + 2Then solve for x:4x = 2x = 2This is the x-coordinate of the vertex. To find the y-coordinate, plug x = 2 back into the original equation:y = -2x2 + 2x + 3y = -8 + 4 + 3y = -1So the vertex is at (2, -1).
2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)
(-2 x2)' = -4 x
Cos-1 2x2-1
4x
Cos-1 2x2-1
Cos-1 2x2-1
d/dx 2x2+3x+7=4x+3
The derivative of 2x2 + 4x + 8 is 4x+4.
d/dx(1/2x) = -1/(2x2)
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
-1
Since this question is in the calculus section, I'm assuming you know how to take the derivative. We know that y = -2x2 + 2x + 3 is a parabola, so it has one vertex, which is a minimum. We can use the first derivative test to find this extreme point.First, take the derivative:y' = -4x + 2Next, set y' equal to zero:0 = -4x + 2Then solve for x:4x = 2x = 2This is the x-coordinate of the vertex. To find the y-coordinate, plug x = 2 back into the original equation:y = -2x2 + 2x + 3y = -8 + 4 + 3y = -1So the vertex is at (2, -1).
2x2 + 4 + 1 = 2x2 + 5 So, the vertex is (0, 5)