How do you find the maximum value of #f(x) = sinx+cosx#?

Answer 1

Please see the explanation.

The x coordinates of extrema can be found by, computing the first derivative, setting that equal to zero, and then solving for x:

Compute the first derivative:

#f'(x) = cos(x) - sin(x)#

Set equal to zero:

#0 = cos(x) - sin(x)#

Solve for x:

#cos(x) = sin(x)#
#1 = sin(x)/cos(x)#
#1 = tan(x)#
#x = tan^-1(1)#
#x = pi/4#
Because the tangent function has a period of #pi#, the value, 1, repeats at every integer multiple of #pi#:
#x = pi/4 + npi# where #n = ...,-3,-2, -1, 0, 1, 2,3,...#

To determine whether this is a maximum, perform the second derivative test, using one of the values:

#f''(x) = -cos(x) - sin(x)#
Evaluate at #pi/4#
#f''(pi/4) = -cos(pi/4) - sin(pi/4) = -sqrt(2)#

The value is a negative, therefore, we have found a maximum.

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Answer 2

To find the maximum value of ( f(x) = \sin x + \cos x ), you first need to find the critical points of the function by taking the derivative and setting it equal to zero. Then, you evaluate the function at these critical points and the endpoints of the interval, if applicable, to determine the maximum value.

Taking the derivative of ( f(x) ) with respect to ( x ) gives:

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

Setting ( f'(x) = 0 ) to find the critical points:

[ \cos x - \sin x = 0 ]

This occurs when ( \cos x = \sin x ), which is true for ( x = \frac{\pi}{4} + k\pi ), where ( k ) is an integer.

Now, evaluate ( f(x) ) at these critical points:

[ f\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} ]

Since ( f(x) ) is periodic, you need to check the function's values on the intervals ( \left[-\frac{\pi}{4}, \frac{\pi}{4}\right] ) and ( \left[\frac{\pi}{4}, \frac{5\pi}{4}\right] ).

[ f\left(-\frac{\pi}{4}\right) = \sin\left(-\frac{\pi}{4}\right) + \cos\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0 ]

[ f\left(\frac{5\pi}{4}\right) = \sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} ]

Comparing these values, the maximum value of ( f(x) ) is ( \sqrt{2} ).

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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.

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.

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.

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.

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