Explanation Given: f ( x ) = sin x 1 − cos x , I = ( 0 , π ] , J = ( 0 , 2 π ) . Calculation: Consider the function f ( x ) = sin x 1 − cos x Now, for critical point f ' ( x ) = 0 d d x ( sin x 1 − cos x ) = 0 ( 1 − cos x ) . cos x − sin x ( sin x ) ( 1 − cos x ) 2 = 0 ( 1 − cos x ) . cos x − sin x ( sin x ) = 0 [Multiplyingboth sides by ( 1 − cos x ) 2 ] 1 − cos 2 x − sin 2 x = 0 [Multiplying both side by x] 1 − ( cos 2 x + sin 2 x ) = 0 1 − 1 = 0 This means a infinite number of critical points are there. Again, lim x → 0 + f ( x ) = lim x → 0 + sin x 1 − cos x [In the form of ∞ ∞ ] lim x → 0 + f ( x ) = lim x → 0 + cos x − ( − sin x ) [UsingL’Hopitals rule] lim x → 0 + f ( x ) = lim x → 0 + cos x sin x lim x → 0 + f ( x ) = lim x → 0 + cot x lim x → 0 + f ( x ) = ∞ Again, lim x → π f ( x ) = lim x → π sin x 1 − cos x [In the form of ∞ ∞ ] lim x → π f ( x ) = lim x → π sin π 1 − cos π lim x → π f ( x ) = 0 1 − ( − 1 ) lim x → π f ( x ) = 0 lim x → ∞ f ( x ) = ∞ The graph of the function with limit I = ( 0 , π ) is shown below:

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Publisher: W. H. Freeman

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Chapter 3.6, Problem 85S

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To find: The global extrema for the given function

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Chapter 3.6, Problem 84S

Chapter 3.6, Problem 86S

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Chapter 3 Solutions

Calculus

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The global extrema for the given function

Solution Summary: The author explains that the function I,f has no global maximum and a global minimum at x=pi .

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