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Absolute min/max

ayahouyee

New member
Nov 4, 2013
12
4)Find the absolute minimum and absolute maximum values of f on the given interval.

a) f(x) = 2cos (x) + sin (2x) on [0, π/2]

b) f(x) = x^2 +16/x on [1, 3]



Thanks again in advance!
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
What have you tried? What method do you know for finding extrema?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
4)Find the absolute minimum and absolute maximum values of f on the given interval.

a) f(x) = 2cos (x) + sin (2x) on [0, π/2]

b) f(x) = x^2 +16/x on [1, 3]



Thanks again in advance!
Hint: Global maxima and minima can occur either at turning points or endpoints.

How do you evaluate the turning points? How do you evaluate the endpoints?
 

ayahouyee

New member
Nov 4, 2013
12
I know that max/min are when f'(x)=0 so

a) f'(x)=-2sin(x)+2cos(2x) and
-2sin(x)+2cos(2x)=0
-sin(x)+cos(2x)=0 and cos(2x)=1-2sin^2(x)
so -sin(x)+1-2sin^2(x)=0
2sin^2(x)+sin(x)-1=0
so (2sinx-1)(sinx+1)=0
so sin(x)=1/2 or sin(x)=-1
giving x=π/6 on [0,π/2]
f"(x)=-2cos(x)-4sin(2x)
= -√3-2√3
f''(x) is <0 so f has a max at x=π/6 and it is an abs max there.

b)f'(x)=2x -16/x² =0 gives
x^3-8=0
gives x=2
f"(x)=2+32/x^3 =6 where x=2
Since f"(2)>0 x=2 is at a min point (2,12)
and on [1,3] this is an abs min point


is this correct? :)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Yes, except that the global maximum or minimum of a function on a closed segment can also lie on the boundary of the segment (see Wikipedia). So you should check the values of the functions at the ends of the intervals and choose the greatest or smallest between those and the values at critical points.