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

Algebra ->  Test -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!

The instantaneous rate of change at x+=+2 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 x+=+3 is negative.
Means: There's a downward slop at x=3
The average rate of change on the interval+0+%3C=x+%3C=4 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 x=2.
or, plot a curve between points:
(0,0) ,(1,3) ,(2,4) ,(3,3) (4,0)

download

you can also find equation using the points above

y=ax%5E2%2Bbx%2Bc..........(0,0)
0=a%2A0%5E2%2Bb%2A0%2Bc
c=0
3=a%2A1%5E2%2Bb%2A1 ..........(1,3)
3=a%2Bb............eq.1

4=a%2A2%5E2%2Bb%2A2+..............(2,4)
4=4a%2B2b......divide by 2
2=2a%2Bb......eq.2
from eq.1 and eq.2 we have system
a%2Bb-3=0
2a%2Bb-2=0
------------------------------
2a%2B2b-6=0...........multiply by 2
2a%2Bb-2=0
------------------------subtract
2a%2B2b-6-2a-b%2B2=0
b-4=0
b=4

3=a%2Bb............eq.1
3=a%2B4
a=3-4
a+=+-1+
y=-x%5E2%2B4x
vertex | (2, 4)

check the average rate of change of f%28x%29=-x%5E2%2B4x on the interval [0,4]
the average rate of change of f%28x%29 on the interval [a,b] is
%28f%28b%29-f%28a%29%29%2F%28b-a%29
we have that a=0, b=4
f%280%29=-0%5E2%2B4%2A0=0
f%284%29=-4%5E2%2B4%2A4=0
thus, %28f%28b%29-f%28a%29%29%2F%28b-a%29=%280-0%29%2F%284-0%29=0

.

Answer by greenestamps(13200) About Me  (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

f%28x%29=a%28x-2%29%5E2

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

f%28x%29+=+-%28x-2%29%5E2

A graph....

graph%28400%2C400%2C-2%2C6%2C-10%2C10%2C-%28x-2%29%5E2%29

(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:

f%28x%29=cos%28%28pi%2F2%29%28x-2%29%29

A graph....

graph%28400%2C400%2C-2%2C6%2C-2%2C2%2Ccos%28%28pi%2F2%29%28x-2%29%29%29

(1) f(0)=f(4)=-1
(2) f'(2)=0
(3) f'(3)<0