we are given two points as
(1.0, 2.0) and (1.0, 5.0)
so,
(x1, y1)= (1.0 , 2.0)
(x2, y2)= (1.0 , 5.0)
x1=1 , y1=2
x2=1 , y2=5
now, we can use distance formula
now, we can plug values
and we get
so, distance between points is 3...........Answer
Help please I need help. :)
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.
