Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Calculus 1 Worksheet # 109 Review Non-Calculator 3 1. f ( x) = x + 1 ⇒ f (1) = 2 while f (−1) = 0 ∴ Neither odd nor even 2. lim x →∞
3. lim
x →−∞
2− x 2x
= lim x →∞
2− x 2x
f ( x) =
5. lim x →0
I.
= lim
( 2 )( 2 )
lim
x →−∞
x →∞
x
1
( 2 )( 2 ) x
x
=
1 1 = = 0 2x 2 ∞ lim
x →−∞
1 = 22 x
lim 2−2 x = 2∞ = ∞
x →−∞
1 1 ⇒ h( x) = f −1 ( x) ⇒ f [ h( x) ] = f ⎡⎣ f −1 ( x) ⎤⎦ = x ⇒ f ' [ h( x) ][ h '( x) ] = 1 ⇒ h '( x) = x f ' [ h( x ) ] 1 f ' [ h(3) ]
4. h '(3) = h '(3) =
=
1
x
1 ⎡1⎤ f '⎢ ⎥ ⎣3⎦
Note: f '( x) =
sin x 0 = 2 x + 3x 0 lim x →1
Note: h(3) = f −1 (3) ⇒ f ( x ) =
x2 −1 = x −1
1 1 1 1 So, 3 = ⇒ x= ∴ h(3) = 3 3 x x
−1 ⎛ 1 ⎞ −1 −1 = −9 ⇒ f '⎜ ⎟ = 2 = 2 x ⎛1⎞ ⎝3⎠ x ⎜ ⎟ ⎝9⎠
∴ L'hopital's rule ⇒
lim x →0
⇒ h '(3) =
1 −9
cos x 1 = 2x + 3 3
lim ( x + 1) = 2 True x →1
6. II. f (1) = 4 True III. f (1) ≠ lim f ( x ) ∴ f ( x ) is discontinuous at x = 1 False x →1
7. A = f + 2 g ⇒ A ' = f '+ 2 g ' ⇒ A '(3) = f '(3) + 2 g '(3) = 4 + 2(−1) = 2
8. B ' = f '• g + f • g ' ⇒ B '(2) = f '(2) • g (2) + f (2) • g '(2) = 3 (1) + 5(−2) = −10 + 3 = −7 9. D =
1 −1 −1 −1 3 1 = g −1 ⇒ D ' = 2 • g ' ⇒ D '(1) = • g '(1) = 2 • −3 = = E 2 g g ( g (1)) 3 9 3
10. H ( x) = 11. K ( x) =
1
f ( x) = [ f ( x) ] 2
⇒ H '( x) =
1 1 1 − •4 = [ f ( x)] 2 • f '( x) ⇒ H '(3) = 2 2 f (3)
2 10
1( 5 ) − 2(−4) 13 f ( x) f '( x) • g ( x) − f ( x) • g '( x) ⇒ K '( x) = ⇒ K '(0) = = 2 g ( x) 52 25 ( g ( x) )
12. M '( x) = f '( g ( x)) • g '( x) ⇒ M '(1) = f '( g (1)) • g '(1) = f '(3) • −3 = 4(−3) = −12 13. P '( x) = f ' ( x3 ) • 3 x 2 ⇒ P '(1) = f '(1) • 3 = 2 • 3 = 6
14.
1 S ( x ) = f −1 ( x ) ⇒ f ⎡⎣ S ( x ) ⎤⎦ = f ⎡⎣ f −1 ( x ) ⎤⎦ = x ⇒ f ' ⎡⎣ S ( x ) ⎤⎦ ⎡⎣ S ' ( x ) ⎤⎦ = 1 ⇒ S ' ( x ) = ⇒ f ' ⎡⎣ S ( x ) ⎤⎦ S '(3) =
1 1 1 Note: S ( 3) = f −1 ( 3) and f (1) = 3 So, f −1 ( 3) = 1 ⇒ S '(3) = = f ' [1] 2 f ' ⎡⎣ S ( 3) ⎤⎦
Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Solutions Calculus 1 Worksheet # 109 Review Non-Calculator 1 1 1 3 − 1 − 1 − 1 1 + 15. y = 2 x 2 − x 2 ⇒ y ' = x 2 + x 2 = 2 4 x 4x x ⎛ ex ⎞ 1 ex −1 ex −1 x x x x y = ln = ln e − ln( e − 1) = x − ln( e − 1) ⇒ y ' = 1 − • e = − = x 16. ⎜ x ⎟ x x x e −1 e −1 e −1 e −1 ⎝ e −1 ⎠
⎛ x⎞ 17. y = tan −1 ⎜ ⎟ ⇒ y ' = ⎝2⎠
18. y ' =
1 2
•
1 1 = • 2 2
1 x2 1+ 4
⎛ x⎞ 1+ ⎜ ⎟ ⎝2⎠ ( e x + e− x )( e x + e− x ) − ( e x − e− x )( e x − e− x )
1 1 1 1 1 x x2 + 1 + x2 2x2 + 1 • 2x = + 2 = = 19. y = ln x x 2 + 1 = ln x + ln ( x 2 + 1) ⇒ y ' = + • 2 2 x 2 ( x + 1) x x + 1 x ( x 2 + 1) x ( x 2 + 1) I g is continuous at x = a because g is differentiable at x = a. True
20. II g is differentiable at x = a because g '(a) exists. True III g is increasing at x = a because g '(a ) > 0. True
