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m147A4sol

# m147A4sol - MATH 147 Assignment#4 Solutions 1 a Find the...

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MATH 147 Assignment #4 Solutions 1) a) Find the derivative of each of the following functions. i) y = cos(3 x 2 ) + arcsin ( x ) ii) f ( x ) = tan( x ) e x iii) f ( x ) = x sin( x ) Solutions i) We have y 0 = - 6 x sin(3 x 2 ) + 1 1 - x 2 . ii) The Quotient Rule shows that f 0 ( x ) = sec 2 ( x ) e x - tan( x ) e x ( e x ) 2 = sec 2 - tan( x ) e x iii) First note that if y = f ( x ) = x sin( x ) , then ln ( y ) = sin ( x ) ln( x ) . Differentiating implicitly gives us that y 0 y = cos( x ) ln ( x ) + sin( x ) x Therefore, y 0 = y (cos( x ) ln ( x ) + sin( x ) x ) = x sin ( x ) (cos( x ) ln ( x ) + sin( x ) x ) Alternatively, we could write x sin( x ) = e sin ( x ) ln ( x ) The Chain Rule shows that 1

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d dx ( x sin( x ) ) = d dx ( e sin( x ) ln( x ) ) = ( e sin ( x ) ln ( x ) ) d dx (sin( x ) ln( x )) = ( e sin ( x ) ln ( x ) )(cos( x ) ln ( x ) + sin( x ) x ) = x sin ( x ) (cos( x ) ln ( x ) + sin( x ) x ) 2) a) Show that f ( x ) = x 2 sin( 1 x 2 ) if x 6 = 0 0 if x = 0 is differentiable every- where but its derivative is not continuous at x = 0. Solution It follows from the standard rules of differentitation that f ( x ) is differentiable whenever x 6 = 0 with derivative f 0 ( x ) = 2 x sin( 1 x 2 ) + ( x 2 cos( 1 x 2 ))( - 2 x 3 ) = 2 x sin( 1 x 2 ) + - 2 cos( 1 x 2 ) x Consider lim h 0 f (0 + h ) - f (0) h = lim h 0 f ( h ) - f (0) h = lim h 0 h 2 sin( 1 h 2 ) h = lim h 0 h sin( 1 h 2 ) Since | sin( 1 h 2 ) |≤ 1 we have - | h |≤ h sin( 1 h 2 ) ≤| h | and since lim h 0 - | h | = 0 = lim h 0 | h | 2
the Squeeze Theorem shows that f 0 (0) = lim h 0 f (0 + h ) - f (0) h = lim h 0 h sin( 1 h 2 ) = 0 . Finally, f 0 ( x ) is not continuous at x = 0. To see, this first note that lim x 0 2 x sin( 1 x 2 ) = 0 . From this it follows that f 0 ( x ) will be continuous at x = 0 if and only if lim x 0 - 2 cos( 1 x 2 ) x = 0 . However, this latter limit does not exist since if x n = 1 2 then x n 0 but lim n →∞ - 2 cos( 1 x 2 n ) x n = lim n →∞ - 2(cos(2 ))( 2 ) = lim n →∞ - 2( 2 ) = -∞ b) Let g ( x ) be such that | g ( x ) |≤ M for all x [ - 1 , 1] . Let h ( x ) = x 2 g ( x ) if x 6 = 0 0 if x = 0 . Show that h ( x ) is differentiable at x = 0 and find h 0 (0) . Solution Observe that h ( x ) - h (0) x = xg ( x ) 3

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for all x [ - 1 , 1] with x 6 = 0. It follows from this that - M | x |≤ h ( x ) - h (0) x M | x | for all x [ - 1 , 1] with x 6 = 0. The Squeeze Theorem clearly shows that h 0 (0) = lim x 0 h ( x ) - h (0) x = 0 Therefore, h ( x ) is differentiable at x = 0 with h 0 (0) = 0. 3) Assume that f ( x ) is continuous on [3 , 5] , f (3) = 2 and that f 0 ( x ) = 1 1+ x 3 on (3 , 5) . Show that 127 63 f (5) 29 14 . Solution Note that f 0 ( x ) is clearly decreasing on [3 , 5]. It follows that f 0 (5) = 1 126 f 0 ( x ) 1 28 = f 0 (3) for any x [3 , 5] The Mean Value Theorem shows that there exists a c (3 , 5) such that f (5) - f (3) 5 - 3 = f 0 ( c ) But then 1 126 f (5) - f (3) 5 - 3 1 28 . This gives us that 1 126 f (5) - 2 2 1 28 and hence that 2 126 + 2 f (5) 2 28 + 2 . That is 127 63 f (5) 29 14 . (Note, these inequalities are actually strict since c (3 , 5) so in fact f 0 (5) = 1 126 < f 0 ( x ) < 1 28 = f 0 (3) for any x (3 , 5). 4
4) Linear Approximation Assume that f ( x ) is differentiable at x = a . The linear approximation to f ( x ) centered at x = a is the function. L a ( x ) = f ( a ) + f 0 ( a )( x - a ) a) Assume that f ( x ) is such that such that f 00 ( x ) exists and is con- tinuous on an inteval I containing x = a. Apply the Mean Value Theorem twice to show that if | f 00 ( x ) |≤ M for every x I , then | f ( x ) - L a ( x ) |≤ M ( x - a ) 2 for each x I.

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