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- Thread starter ayahouyee
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- Jan 30, 2012

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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.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!

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

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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

- 2,492