How do you find max's or min's? I assume this is from a Caculus course.Originally Posted by yiyayiyayo
So
Now find f'(x) and set it equal to zero. There's your start.
" If f(x)=(sinx+3)(cosx+2)/2
Then f(x)max=?
and what about f(x)min=? "
Is that really f(x) = (sin(x) +3)(cos(x) +2)/2 ?
Not f(x) = (sin(x+3))(cos(x+2))/2 ?
Let us say,
f(x) = (sin(x) +3)(cos(x) +2)/2 -------------(i)
One way to find max f(x) and min f(x) is to graph the function and then analyze the graph. Eyeball the possible max or min f(x), then solve for them by iteration.
As given, f(x) is difficult to graph, so let us convert it into one that can be graphed easily.
f(x) = (sin(x) +3)(cos(x) +2)/2
2f(x) = (sin(x) +3)(cos(x) +2)
2f(X) = (sinX +3)(cosX +2)
2f(X) = sinX*cosX +2*sinX +3*cosX +3*2
2f(X) = sinXcosX +2sinX +3cosX +6
2f(X) = (1/2)sin(2X) +2sinX +3cosX +6 ------------(ii)
Now that is easier to graph.
On the same x,y coordinates setup, plot or graph :
y = (1/2)sin(2X) ---------period is 360/2 = 180 degrees.
y = 2sinX ----period is 360 deg.
y = 3cosX ---period is 360 deg.
y = 6 ----a horizontal line only.
Do that on paper, graphing calculator, or wherever.
I sketced them on paper here.
I saw/thought that the max f(x) is between X=0 and x=pi/4, or between X=0deg and x=45deg.
So I tested X=35 deg on the original Eq.(i).
Then by iteration I tested X=36, X=37, X=38. It seemed that max f(x) is between x=37deg and x=38deg.
By further iteration, I found max f(X) is f(37.7deg) = 5.040289.
For the min f(x), I thought it was between X=pi and X=5pi/4, or between X=180deg and X=225deg.
By iteration I found min f(X) is f(202.3deg) = 1.408267.