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- #1
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Absolute min/max
- Thread starter ayahouyee
- Start date
- Jan 30, 2012
- 2,528
What have you tried? What method do you know for finding extrema?
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?
- Thread starter
- #4
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?
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?
- Jan 30, 2012
- 2,528
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.