How do you find the critical numbers for #f(x) = x + 2sinx# to determine the maximum and minimum?

Answer 1

William Abdalla

The critical points of a function #f(x)# are the #x# that make #f'(x)=0#

We calculate the derivative #f'(x)=1+2cos(x)#, and now we need to find where #f'(x)=1+2cos(x)=0#. But that means:

#-1=2cos(x)#, and then #cos(x) = -1/2#. Between #0# and #2pi# these points are:

#x=4/6pi=2/3pi# and #8/6pi=4/3pi#,

and all the congruents are:

#x=2/3pi+K*2pi# and #4/3pi+K*2pi# where #K# is any integer number (positive or negative)

Now the second derivative of #f(x)# is:

#f''(x) = -2sin(x)# which is negative in the first set of points and positive in the second set. So the points:

#x=2/3pi+K*2pi# are all local maxima, and the points #4/3pi+K*2pi# are all local minima

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Answer 2

Aurora Alvarado

To find the critical numbers of ( f(x) = x + 2\sin(x) ), where the maximum and minimum may occur, follow these steps:

Find the derivative of ( f(x) ) with respect to ( x ).
Set the derivative equal to zero and solve for ( x ).
Determine whether the critical numbers correspond to maximum, minimum, or neither by using the first or second derivative test.

Let's find the critical numbers:

Find the derivative of ( f(x) ): [ f'(x) = 1 + 2\cos(x) ]
Set the derivative equal to zero and solve for ( x ): [ 1 + 2\cos(x) = 0 ] [ \cos(x) = -\frac{1}{2} ]

To find the solutions for ( x ) where ( \cos(x) = -\frac{1}{2} ), we look at where the cosine function equals ( -\frac{1}{2} ). These solutions occur at ( x = \frac{2\pi}{3} + 2\pi n ) and ( x = \frac{4\pi}{3} + 2\pi n ), where ( n ) is an integer.

Determine the nature of critical points using the first derivative test or second derivative test. If ( f''(x) > 0 ) at a critical point, it corresponds to a local minimum; if ( f''(x) < 0 ), it corresponds to a local maximum. If the second derivative test is inconclusive, you can use the first derivative test.

[ f''(x) = -2\sin(x) ]

At ( x = \frac{2\pi}{3} + 2\pi n ): [ f''\left(\frac{2\pi}{3} + 2\pi n\right) = -2\sin\left(\frac{2\pi}{3} + 2\pi n\right) = -\sqrt{3} ] Since ( f''\left(\frac{2\pi}{3} + 2\pi n\right) < 0 ), it corresponds to a local maximum.

At ( x = \frac{4\pi}{3} + 2\pi n ): [ f''\left(\frac{4\pi}{3} + 2\pi n\right) = -2\sin\left(\frac{4\pi}{3} + 2\pi n\right) = \sqrt{3} ] Since ( f''\left(\frac{4\pi}{3} + 2\pi n\right) < 0 ), it corresponds to a local maximum.

Therefore, the critical numbers for ( f(x) = x + 2\sin(x) ) occur at ( x = \frac{2\pi}{3} + 2\pi n ) and ( x = \frac{4\pi}{3} + 2\pi n ), where ( n ) is an integer. At these critical points, ( f(x) ) has local maximums.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.