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
Related questions
Question
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 question
Image 2: My work
Transcribed Image Text:Consider the following piecewise function: f(x) =
XCOLE x<0
0≤x≤4
a. Use the first principles definition of a derivative to determine f'(x) when 0 < x < 4
b. Determine f"(3)
c. Determine the equation of the tangent to f(x) when x = - =
d. Determine f'(x) when x > 4
e. Is f(x) continuous when x = 4? Use the formal definition of continuity to argue your answer...
Transcribed Image Text:c) f(x) = e
XCOSX
d
clx
4
-7 Let u = Xcos x
X LO
= a (xcosx)
=xd (cosx) + cos x d (x)
=-xsinx + cos x
when x = -7/2
Now f(x) = e
-7 Differentiate with respect to 'x'
d
f'(x) = a (eº)
d
du (e) du (using chain irule)
(using product Rule)
seo (cosxxsinx)
= excosx (Cosx-xsinx)
f(-1/2) = e
풋
=-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 = - 11/2
and f(-171/2) = 1 is given by
(y-1) = f'(-1/2)(x + 1)
=> (g-1) = = = (x + 1)
TX
=> y= - 1² - 1 ² + 1
2
Thus the equation of the
tangent at x = - 71/2 15
y = 1-1/2 (x + 7/2)
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