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

  • MHB
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In summary: Also, you should mention the values of f at the critical points as well as the values of x at the end points in order to determine which of the values is the global maximum or minimum. In summary, the absolute minimum value of f on [0, π/2] is -1 and it occurs at x=π/6, while the absolute maximum value is 1 and it occurs at x=0. On [1, 3], the absolute minimum value of f is 12 and it occurs at x=2, while the absolute maximum value is 25 and it occurs at x=3.
  • #1
ayahouyee
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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!
 
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  • #2
What have you tried? What method do you know for finding extrema?
 
  • #3
ayahouyee said:
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?
 
  • #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 pointis this correct? :)
 
  • #5
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.
 

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

What is the process for finding the minimum and maximum values of a function?

The process for finding the minimum and maximum values of a function involves taking the derivative of the function, setting it equal to zero, and solving for the critical points. Then, we plug these critical points back into the original function to determine the corresponding y-values. The smallest y-value is the minimum and the largest y-value is the maximum.

What is the difference between a local minimum/maximum and a global minimum/maximum?

A local minimum/maximum is a point on the graph of a function that is the smallest/largest value within a small interval, but it may not be the smallest/largest value of the entire function. A global minimum/maximum is the smallest/largest value of the entire function.

Can a function have more than one minimum/maximum?

Yes, a function can have multiple local minimums/maximums, but it can only have one global minimum/maximum.

How do we determine if a critical point is a minimum or maximum?

To determine if a critical point is a minimum or maximum, we can use the second derivative test. If the second derivative at the critical point is positive, the critical point is a minimum. If the second derivative is negative, the critical point is a maximum. If the second derivative is zero, the test is inconclusive.

Is there a way to find the minimum and maximum values without using calculus?

Yes, we can also use the method of graphing or creating a table of values to estimate the minimum and maximum values of a function. However, this method may not always give an accurate result and it is recommended to use calculus for a more precise answer.

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