What is the greatest common factor of 14 56 63?
1 answer:
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9514 1404 393
Answer:
- maximum: 2
- minimum: -6
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
f(0) = 2
f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.
Since you're only asked the ordered pair of D'', it's much easier just to plot and reflect point D twice than to do that for all four points!
Remember that reflecting points is like putting a mirror at the line of reflection or flipping that point over at that line. The reflected point should be the same distance from the line of reflection as the original point.
1) Reflect D over the x-axis to get D'.
D is at (4,1). Draw a line that is perpendicular to the line of reflection and goes through D. D is as far from the line of reflection as D' should be on its other side (both are on that perpendicular line). Since D is 1 unit above the x-axis, that means D' is 1 unit below at (4, -1). See picture 1.
2) Reflect D' over <span>y=x+1 to get D''.
D' is at (4, -1). Draw </span>y=x+1 and the line perpendicular to it going through D''. D'' is the same distance from the line of reflection on the other side. See picture 2. D'' is at (-2, 5).
Answer: D'' is at (-2, 5)
Remember that reflecting points is like putting a mirror at the line of reflection or flipping that point over at that line. The reflected point should be the same distance from the line of reflection as the original point.
1) Reflect D over the x-axis to get D'.
D is at (4,1). Draw a line that is perpendicular to the line of reflection and goes through D. D is as far from the line of reflection as D' should be on its other side (both are on that perpendicular line). Since D is 1 unit above the x-axis, that means D' is 1 unit below at (4, -1). See picture 1.
2) Reflect D' over <span>y=x+1 to get D''.
D' is at (4, -1). Draw </span>y=x+1 and the line perpendicular to it going through D''. D'' is the same distance from the line of reflection on the other side. See picture 2. D'' is at (-2, 5).
Answer: D'' is at (-2, 5)