Question 1183710
<br>
Average rate of change on [0,4] zero means f(0)=f(4).<br>
Instantaneous rate of change zero at x=2 means there is a local maximum or minimum at x=2.<br>
Those two conditions together are easily satisfied by a function of the form<br>
{{{f(x)=a(x-2)^2}}}<br>
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.<br>
So one simple function satisfying the given conditions is<br>
{{{f(x) = -(x-2)^2}}}<br>
A graph....<br>
{{{graph(400,400,-2,6,-10,10,-(x-2)^2)}}}<br>
(1) f(0)=f(4)=-4
(2) f'(2)=0
(3) f'(3)<0<br>
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<br>
This function satisfies those conditions:<br>
{{{f(x)=cos((pi/2)(x-2))}}}<br>
A graph....<br>
{{{graph(400,400,-2,6,-2,2,cos((pi/2)(x-2)))}}}<br>
(1) f(0)=f(4)=-1
(2) f'(2)=0
(3) f'(3)<0<br>