x b f parenrightbigg f x 6 a F prime x 1 1 x 2 2 2

X b f parenrightbigg f x 6 a f prime x 1 1 x 2 2 2

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Unformatted text preview: parenleftbiggintegraldisplay x b f parenrightbigg =- f ( x ). 6. (a) F prime ( x ) = 1 [1 + ( x 2 ) 2 ] 2 (2 x ) = 2 x (1 + x 4 ) 2 (b) F prime ( x ) = √ 1 + x 2- √ 1 + x 2 = 0. Note: integraldisplay x radicalbig 1 + t 2 dt + integraldisplay 2 x radicalbig 1 + t 2 dt = integraldisplay 2 radicalbig 1 + t 2 dt , a constant. Therefore, F prime ( x ) = 0. (c) F prime ( x ) = sin(3 x ) 2 (3) = 3 sin 9 x 2 . (d) F prime ( x ) = cos √ x 4 (4 x 3 )- cos √ x 2 (2 x ) = 4 x 3 cos x 2- 2 x cos x . 7. F ( x ) = integraldisplay x xe t 2 dt = x integraldisplay x e t 2 dt . F prime ( x ) = xe x 2 + integraldisplay x e t 2 dt F primeprime ( x ) = e x 2 + 2 x 2 e x 2 + e x 2 = 2 e x 2 + 2 x 2 e x 2 8. f ( t ) = t, ≤ t ≤ 2 3 , 2 < t ≤ 4 . F ( x ) = integraldisplay x f ( t ) dt (a) For 0 ≤ x ≤ 2 , F ( x ) = integraldisplay x tdt = 1 2 t 2 bracketrightbig x = 1 2 x 2 . For 2 < x ≤ 4 F ( x ) = integraldisplay x f ( t ) dt = integraldisplay 2 f ( t ) dt + integraldisplay x 2 f ( t ) dt = 2 + integraldisplay x 2 3 dt = 2 + 3 t bracketrightbig x 2 = 2 + 3 x- 6 = 3 x- 4 . Therefore, F ( x ) = 1 2 x 2 , ≤ x ≤ 2 3 x- 4 , 2 < x ≤ 4 3 (c) F prime ( x ) = x, ≤ x < 2 3 , 2 < x ≤ 4 ; F is not differentiable at x = 2. 10. lim x → integraltext x √ 9 + t 2 dt x has the form 0 / 0. Therefore, L’Hospital’s Rule applies. lim x → √ 9 + x 2 1 = 3. Therefore, lim x → integraltext x √ 9 + t 2 dt x = 3. 11. Given that f ( x ) negationslash = 0 for all x , and [ f ( x )] 2 = 2 integraldisplay x f ( t ) dt . Differentiate this equation: 2 f ( x ) f prime ( x ) = 2 f ( x ) ⇒ f prime ( x ) = 1 ⇒ f ( x ) = x + C. Since f (0) = 0 , C = 0 and f ( x ) = x . 12. For any x ∈ [ a, b ], integraldisplay x a f ( t ) dt + integraldisplay b x f ( t ) dt = integraldisplay b a f ( t ) dt . Since we are given that integraldisplay b x f ( t ) dt = integraldisplay x a f ( t ) dt , we have 2 integraldisplay x a f ( t ) dt = integraldisplay b a f ( t ) dt. Of course, integraldisplay b a f ( t ) dt = k , a number, so calculating the derivative of this equation, we get 2 f ( x ) = 0 and f ( x ) = 0. Since x was “any” point in [ a, b ], f ( x ) = 0 for all x ∈ [ a, b ]....
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