Please help me with this. Please check if I've gotten the write answer for each questions. And if I have shown the correct steps. Just check question c)Just write the answers that are correct (e.g. a) correct, b) incorrect )Image 1: The questionImage 2: My work Consider the following piecewise function: f(x) =XCOLE x<00≤x≤4a. Use the first principles definition of a derivative to determine f'(x) when 0 < x < 4b. Determine f"(3)c. Determine the equation of the tangent to f(x) when x = - =d. Determine f'(x) when x > 4e. Is f(x) continuous when x = 4? Use the formal definition of continuity to argue your answer... c) f(x) = eXCOSXdclx4-7 Let u = Xcos xX LO= a (xcosx)=xd (cosx) + cos x d (x)=-xsinx + cos xwhen x = -7/2Now f(x) = e-7 Differentiate with respect to 'x'df'(x) = a (eº)ddu (e) du (using chain irule)(using product Rule)seo (cosxxsinx)= excosx (Cosx-xsinx)f(-1/2) = e풋=-22- cos (-1/2)2-T1/2 (0)f'( - ²) = 1 [00³ (-²) - (-) sin(-)= 1 (0-1)Equation of the tangent at the point x = - 11/2and f(-171/2) = 1 is given by(y-1) = f'(-1/2)(x + 1)=> (g-1) = = = (x + 1)TX=> y= - 1² - 1 ² + 12Thus the equation of thetangent at x = - 71/2 15y = 1-1/2 (x + 7/2)

Answered: Image /qna-images/answer/9233e707-c34f-4948-8068-4e00a7c562e1.jpg

c) f(x) = e XCOSX ixco -7 Let u=XCOS X d = a (xcosx) clx xd (cosx) + cos x cos X d (x) E-xsinx & COS X Now f(x) = eu -7 Differentiate with respect to 'x' f(x) = (e) when x = - 11/2 (using product Rule) d (e) du (using chain itule) du dx co (cosxxsinx) =excosx (cosxxsinx) f(-1/2)= cos(- 11/2) -T1/2 (0) 6=e8=1 f'(-1) = 1 [cos(-¹)-(-3)sin (-1) 1 (0-1) Equation of the tangent at the point x = - 11/2 and f(-11/2) = is given by (y-1)= f'(-1/2)(x+) => (g-1) = (x+1) => y= -2 -²+1 Thus the equation of the tangent at x = -7/2 15 y = 1-1/2 (x + π/2)

c) f(x) = e XCOSX ixco -7 Let u=XCOS X d = a (xcosx) clx xd (cosx) + cos x cos X d (x) E-xsinx & COS X Now f(x) = eu -7 Differentiate with respect to 'x' f(x) = (e) when x = - 11/2 (using product Rule) d (e) du (using chain itule) du dx co (cosxxsinx) =excosx (cosxxsinx) f(-1/2)= cos(- 11/2) -T1/2 (0) 6=e8=1 f'(-1) = 1 [cos(-¹)-(-3)sin (-1) 1 (0-1) Equation of the tangent at the point x = - 11/2 and f(-11/2) = is given by (y-1)= f'(-1/2)(x+) => (g-1) = (x+1) => y= -2 -²+1 Thus the equation of the tangent at x = -7/2 15 y = 1-1/2 (x + π/2)

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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