MHB Find Min/Max of f: 2cos(x)+sin(2x) & x^2+16/x

[ayahouyee]

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]

What have you tried? What method do you know for finding extrema?

 

[Prove It]

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]

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 pointis this correct? :)

 

[Evgeny.Makarov]

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.