SOLUTION: 11) Create a graph of a function given the following information:
• The instantaneous rate of change at x = 2
is zero.
• The instantaneous rate of change at x = 3
is negati
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Question 1183710: 11) Create a graph of a function given the following information:
• The instantaneous rate of change at x = 2
is zero.
• The instantaneous rate of change at x = 3
is negative.
• The average rate of change on the interval 0 <= x <= 4 is zero.
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
The instantaneous rate of change at is zero.
Means: A peak trough or inflection. The curve levels off there; the tangent is parallel to the x-axis.
The instantaneous rate of change at is negative.
Means: There's a downward slop at
The average rate of change on the interval is zero.
Means: It's symmetric over the interval - for every up there is a down, and you end as high as you start.
Thus: It's a hill. Plot a parabola with a peak at .
or, plot a curve between points:
(,) ,(,) ,(,) ,(,) (,)
you can also find equation using the points above
..........(,)
..........(,)
............eq.1
..............(,)
......divide by
......eq.2
from eq.1 and eq.2 we have system
------------------------------
...........multiply by
------------------------subtract
............eq.1
vertex | (, )
check the average rate of change of on the interval [,]
the average rate of change of on the interval [,] is
we have that ,
thus,
.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
Average rate of change on [0,4] zero means f(0)=f(4).
Instantaneous rate of change zero at x=2 means there is a local maximum or minimum at x=2.
Those two conditions together are easily satisfied by a function of the form
To have the instantaneous rate of change negative at x=3, we simply need to make the parabola open downward, which means a is negative.
So one simple function satisfying the given conditions is
A graph....
(1) f(0)=f(4)=-4
(2) f'(2)=0
(3) f'(3)<0
A sinusoidal function can also be found that satisfies the given conditions:
f(0)=f(4) means the period of the function can be 4
f'(2)=0 and f'(3)<0 means we want a local maximum at x=2
This function satisfies those conditions:
A graph....
(1) f(0)=f(4)=-1
(2) f'(2)=0
(3) f'(3)<0
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