Midterm Practice 1 Solutions PDF

Preview

Summary

Download Midterm Practice 1 Solutions PDF

Description

Math 114 — Solutions to Midterm Practice Problems 1. Find y′ . Do not simplify. (a) y = (1 − 3x + ex )−3 (x − cos2 x)7 Solution. y′ = −3(1 − 3x + ex )−4 (−3 + ex )(x − cos2 x)7

+ (1 − 3x + ex )−3 · 7(x − cos2 x)6 [1 − 2(cos x)(− sin x)]

(b) y =



3 − x3 2x2 + 1

 32

Solution. 2 y = 3 ′



3 − x3 2x2 + 1

− 31 

−3x2 (2x2 + 1) − (3 − x3 )(4x) (2x2 + 1)2



√ 3 (c) y = sin4 (tan 1 − ex ) Solution. y′ = 4[sin3 (tan

(d) y =

p

3

1 − ex )][cos(tan

p

1 − ex )](sec2 3

p

3

1 − ex )



 3 1 3 1 (1 − ex )− 2 (−ex )(3x2 ) 2

tan2 x 1 − sec2 x

Solution. y′ = (or note that y =

tan2 x 1−sec2 x

2 tan x sec2 x(1 − sec2 x) − tan2 x(−2 sec x)(sec x tan x) (1 − sec2 x)2 =

tan2 x − tan2 x

= −1, so y ′ = 0; the above will simplify to 0).

√ (e) y = csc2 (x2 − sin 1 + x) Solution. √ y′ = 2(csc(x2 − sin 1 + x))    √ √ √   1 1 (1 + x)− 2 · − csc(x2 − sin 1 + x) cot(x2 − sin 1 + x) 2x − (cos 1 + x) 2 (f) y =

p

1 − sin(x2 − 2e5x ).

Solution. y′ = 2

(g) y = tan(esec

(1−x3 )

)

− 1 1 1 − sin(x2 − 2e5x ) 2 (− cos(x2 − 2e5x ))(2x − 2e5x · 5). 2

Page 2 Solution. 2

y′ = [sec2 (esec

(1−x3 )

2

)][esec

(1−x3 )

][2 sec(1 − x3 )(sec(1 − x3 ) tan(1 − x3 ))(−3x2 )]

2. Evaluate the limits. (a) lim

x→1

x2 + x − 2 x2 − 1

Solution. lim

x→1

(b) lim

x→2

x2 + x − 2 3 x+2 (x + 2)(x − 1) = . = lim = lim x→1 (x − 1)(x + 1) x→1 x + 1 x2 − 1 2

√ √ x + 2 − 2x x2 − 2x

Solution. lim

√

x→2

(c) lim

x→2

√ √ √ √ √ ( x + 2 − 2x)( x + 2 + 2x) x + 2 − 2x x + 2 − 2x √ √ √ = lim √ = lim x→2 x→2 x2 − 2x x(x − 2)( x + 2 + 2x) x(x − 2)( x + 2 + 2x) −1 2−x −1 . = lim √ √ √ = lim √ = x→2 x(x − 2)( x + 2 + 2x) x→2 x( x + 2 + 2x) 8

|2 − x| x2 − 4

Solution. 1 2−x −1 |2 − x| =− = lim = lim x2 − 4 x→2− (x − 2)(x + 2) x→2− x + 2 4 1 1 −(2 − x) |2 − x| lim 2 = . = lim+ = lim + + x + 2 ( x − 2)( x + 2) x − 4 4 x→2 x→2 x→2 lim

x→2−

Since the left hand limit and right hand limit are not the same, limx→2 |2 − x|/(x2 − 4) does not exist. x3 − 1 (d) lim √ x→1 x−1 Solution. √ √ (x3 − 1)( x + 1) (x − 1)(x2 + x + 1)( x + 1) x3 − 1 √ = 3(2) = 6. = lim lim √ = lim √ x→1 x→1 x−1 x − 1 x→1 ( x − 1)( x + 1) Or

√ √ x3 − 1 (x − 1)(x2 + x + 1) ( x − 1)( x + 1)(x2 + x + 1) lim √ √ √ = lim = lim = 6. x→1 x→1 x − 1 x→1 x−1 x−1

(e) lim

θ→0

1 − cos 2θ sin2 θ cos θ

Solution. 2 2 sin2 θ 2 1 − cos 2θ = = 2. = lim = lim 2 2 θ→0 sin θ cos θ θ→0 sin θ cos θ θ→0 cos θ 1 lim

Page 3 (Or multiply the numerator and denominator by 1 + cos 2θ but this method takes much longer.) (f) lim

1 x+1

x→1

1 2

−

x−1

Solution. lim

x→1

(g) lim

x→0

1−

1 x+1

−

x−1

1 2

2 − (x + 1) 1−x = lim 2(x + 1)(x − 1) x→1 2(x + 1)(x − 1) 1 −1 −(x − 1) =− . = lim = lim x→1 2(x + 1) x→1 2(x + 1)(x − 1) 4 = lim

x→1

x √ x+1

Solution. √ √ √ x x(1 + x + 1) 1+ 1+x x(1 + x + 1) √ = lim √ √ = lim lim = lim = −2. x→0 x→0 x→0 1 − x + 1 x→0 (1 − x + 1)(1 + x + 1) 1−x−1 −1

(h) lim

x→0

tan 3x 4x

Solution. lim

x→0

      3 1 1 3 3x sin 3x sin 3x 3 6 x sin 3x tan 3x · = lim = lim = 1· ·1 = . = lim x→0 x→0 x→0 4x cos 3x 3x 4 4 4x 3x 4 6 x cos 3x 4x cos 3x

(i) lim+ x→3

x2 − 5x + 6 |3 − x| Since |3 − x| = −(3 − x) when x > 3, we get

Solution. lim

x→3+

(j) lim x cos x→0−

x2 − 5x + 6 (x − 3)(x − 2) (x − 3)(x − 2) = lim+ = lim+ = lim+ (x − 2) = 1. x→3 |3 − x| −(3 − x) x−3 x→3 x→3

1 x

Solution. Note that −1 ≤ cos 1x ≤ 1. Let x < 0. Then x ≤ x cos 1x ≤ −x and limx→0− (−x) = limx→0− x = 0. Then by the Squeeze Theorem, limx→0− x cos x1 = 0.

(k)

lim

x→−∞

1 − x − 3x2 5 + x3

Solution. lim

x→−∞

1 − x − 3x2 = lim x→−∞ 5 + x3

1 x3

−

5 x3

1 x2

− x3

+1

=

0 0−0−0 = = 0. 0+1 1

Page 4 (l) lim− x→3

x(x − 1) (x + 1)(x − 3) As x → 3− (x < 3),

Solution.

x → 3, ∴ ∴

(m)

lim

x→−∞

x − 1 → 2,

x + 1 → 4,

x(x − 1) is a large negative number (x + 1)(x − 3) x(x − 1) = −∞. lim − x→3 (x + 1)(x − 3)

√ 3x2 − x + 5 5x + 3

Solution.

lim

x→−∞

Since √

√ x2 = −x, x > 0,

3x2

−x+5 = lim x→−∞ 5x + 3

q

3x2 −x+5 x2 5x+3 −x

= lim

x→−∞

3. Find an equation of the tangent line to the curve y = Solution.

and x − 3 → 0 negatively

q

3−

1 x

+

5 x2

3 x

−5 −

√ √ 3 3−0+0 . = − = 5 −5 − 0

x at x = −2. 1−x

The slope of the tangent line to the curve is given by y′ =

1 1−x+x 1 − x − x(−1) = . = (1 − x)2 (1 − x)2 (1 − x)2

So at x = −2, the slope is y ′ (−2) = 1/(1 − (−2))2 = 1/(1 + 2)2 = 1/9 Also, y = −2/(1 − (−2)) = −2/3. So the tangent line is 1 2 y + = (x + 2). 3 9 4. Find an equation(s) of the tangent line(s) to the curve y = the tangent line(s).

1 through (3, −1). Sketch the graph and x

Solution. The slope of the tangent line to the curve is y ′ = −1/x2 . Let (a, 1/a) be the point of tangency. Then the slope of tangent at that point is y ′ (a) = −1/a2 . Also, using the points (a, 1/a) and 1/a+1 (3, −1) and the slope formula the slope of the tangent line is: m = a−3 . Then we can equate the two slopes: 1 a

+1 1 1+a 1 = − 2 =⇒ a2 (1 + a) = −a(a − 3) =⇒ a + a2 = −a + 3 = − 2 =⇒ a a(a − 3) a a−3 =⇒ a2 + 2a − 3 = 0 =⇒ (a + 3)(a − 1) = 0 =⇒ a = −3, 1.

Finding the y-values for the above x-values, the points of tangency are (−3, − 13 ) and (1, 1). Then the tangent lines are: at (−3, −1/3) with slope y ′ (−3) = − at (1, 1) with slop y ′ (1) = −1

=⇒

1 9

1 1 = − (x + 3) 3 9 y − 1 = −(x − 1). =⇒

y+

Page 5 y

(a, 1/a)=(1,1)

x

0 (-3,1)

(3,-1)

5. For the following use the definition of the derivative of f (x).

(a) Find f ′ (x) given f (x) =

x+4 . x−5

Solution. x+h+4 − x+4 (x + h + 4)(x − 5) − (x + 4)(x + h − 5) f (x + h) − f (x) = lim x+h−5 x−5 = lim h→0 h→0 h→0 h(x + h − 5)(x − 5) h h x2 − 5x + xh − 5h + 4x − 20 − x2 − xh + 5x − 4x − 4h + 20 = lim h→0 h(x + h − 5)(x − 5) −9 −9h −9 = lim . = = lim (x − 5)2 h→0 h(x + h − 5)(x − 5) h→0 (x + h − 5)(x − 5)

f ′ (x) = lim

1 (b) Find f ′ (x) given f (x) = √ . 3x + 1 Solution.

′

f (x) = = = = = =

p √ √ 1 − √3x1+1 3x + 1 − 3(x + h) + 1 3(x+h)+1 f (x + h) − f (x) = lim p = lim lim √ h→0 h 3(x + h) + 1 3x + 1 h→0 h→0 h h p p √ √ ( 3x + 1 − 3(x + h) + 1)( 3x + 1 + 3(x + h) + 1) p p lim √ √ h→0 h 3(x + h) + 1 3x + 1( 3x + 1 + 3(x + h) + 1) 3x + 1 − 3x − 3h − 1 lim p p √ √ h→0 h 3(x + h) + 1 3x + 1( 3x + 1 + 3(x + h) + 1) −3h p lim p √ √ h→0 h 3(x + h) + 1 3x + 1( 3x + 1 + 3(x + h) + 1) −3 p lim p √ √ h→0 3(x + h) + 1 3x + 1( 3x + 1 + 3(x + h) + 1) −3 3 √ √ √ . =− 2(3x + 1)3/2 3x + 1 3x + 1(2 3x + 1)

(c) Find f ′ (0) given f (x) =

√ 1 − x2 .

Page 6 Solution. f (x) − f (0) = lim f (0) = lim x→0 x→0 x ′

= lim

x→0

√

√ √ 1 − x2 − 1 ( 1 − x2 − 1)( 1 − x2 + 1) √ = lim x→0 x x( 1 − x2 + 1)

0 −x −x2 1 − x2 − 1 = = 0. √ = lim √ = lim √ 2 x( 1 − x2 + 1) x→0 x( 1 − x2 + 1) x→0 1 − x2 + 1

6. Show that the equation x4 + 4x − 1 = 0 has at least one root in the interval (0, 1). Solution. Let f (x) = x4 + 4x − 1. This function is continuous on [0, 1] since it is a polynomial (it’s continuous everywhere). Also f (0) = −1 < 0 and f (1) = 1 + 4 − 1 = 4 > 0 so that f (0) < 0 < f (1). Then by the Intermediate Value Theorem there exists a number c ∈ (0, 1) such that f (c) = 0, i.e., the equation x4 + 4x − 1 = 0 has a root in (0, 1). 7. Determine whether or not the function f is continuous at 4.  if x = 4  3, f (x) = . x2 − 2x − 8  , if x 6= 4 x−4 Solution.

f is continuous at x = 4 if limx→4 f (x) = f (4). Now lim f (x) = lim

x→4

x→4

(x − 4)(x + 2) x2 − 2x − 8 = lim = lim (x + 2) = 6. x→4 x→4 x−4 x−4

But f (4) = 3 6= 6. So f is discontinuous at x = 4. 8. Determine whether or not the function f is continuous on its domain. √ x 3 Solution. The domain of f is (−∞, ∞). It is easy to see that f is continuous on (−∞, 0) ∪ (0, 3) ∪ (3, ∞). We now check to see if f is continuous at x = 0 and x = 3. We need to check: limx→0 f (x) = f (0) and limx→3 f (x) = f (3)? lim f (x) = lim − −

x→0

x→0

√ −x = 0

and

lim f (x) = lim (3 − x) = 3. +

x→0+

x→0

So limx→0 f (x) does not exist, i.e., f is discontinuous at x = 0. This is enough to say f is not continuous on its domain. But just to check at x = 3, lim f (x) = lim− (3 − x) = 0

x→3−

x→3

and

lim f (x) = lim+ (x − 3)2 = 0,

x→3+

x→3

so limx→3 f (x) = 0 and f (3) = 3 − 3 = 0 = limx→3 f (x) so f is continuous at x = 3. So f is continuous only on (−∞, 0) ∪ (0, ∞). 9. Find the values of b and k so that the function is differentiable at x = 2.  bx, x 0, i.e., x < 3, i.e., (−∞, 3). We find the intersection of these sets so that the domain of f is (−∞, 2) ∪ (2, 3). (b) To√find√the domain of f , we set 2 − x2 ≥ 0, i.e., x2 ≤ 2 =⇒ |x| ≤ [− 2, 2].

√ √ √ 2. Hence, − 2 ≤ x ≤ 2, i.e.,

√ x2 − 1 (c) f is defined when √ > 0. Since 3 − x2 > 0, x2 − 1 > 0, i.e., x2 > 1 ⇐⇒ |x| > 1. Hence, 2 3−x x < −1 or x > 1 i.e., (−∞, −1) ∪ (1, ∞). √ √ √ Also, we must have 3 − x2 > 0, i.e., x2 < 3 ⇐⇒ |x| < 3 ⇐⇒ − 3 < x < 3. Therefore, f is defined in the intersection of these two sets: √ √ √ √ {(−∞, −1) ∪ (1, ∞)} ∩ (− 3, 3) = (− 3, −1) ∪ (1, 3). (d) First we find the points at which ln(x + 1) = 5. We get x + 1 = e 5 =⇒ x = e5 − 1. So we want x 6= e5 − 1. As well, we need x + 1 > 0, i.e., x > −1. Putting these results together, we see that the domain of f is (−1, e5 − 1) ∪ (e5 − 1, ∞). 11. Let h(x) = f (g(sin 4x)) where f and g are differentiable functions. (a) Find h′ (x) Solution.

Using the Chain Rule, h′ (x) = [f ′ (g(sin 4x))][g ′ (sin 4x)][cos 4x](4).

Page 8 (b) Find h′ (0) given that g(0) = 1, g ′ (0) = −1 and f ′ (1) = 2. Solution.

Using the result of part (a),

h′ (0) = 4[f ′ (g(sin 0))][g ′ (sin 0)][cos 0] = 4f ′ (g(0)) · g ′ (0) · 1 = 4(f ′ (1))(g ′ (0)) = 4(2)(−1) = −8. 12. Solve the following inequalities: (a) x2 > 2x + 6 Solution.

(b) |3 − 7x| ≥ 4

(c)

1 2x + 6 =⇒ x2 − 2x − 6 > 0. Completing the square, we obtain (x − 1)2 − 7 > 0 =⇒ (x − 1)2 > 7 =⇒ |x − 1| > √ √ =⇒ x − 1 < − 7 or x − 1 > 7 √ √ =⇒ x < 1 − 7 or x > 1 + 7.

In the interval form: (−∞, 1 −

√ 7

√ √ 7) ∪ (1 + 7, ∞).

(b) |3 − 7x| ≥ 4 =⇒ 3 − 7x ≤ −4 or 3 − 7x ≥ 4. Solving these inequalities, we get x ≥ 1 or x ≤ −1/7, i.e., (−∞, −1/7] ∪ [1, ∞). 1 < 3. Since the denominator contains a variable, we need to consider 2 cases. x If x > 0, then 1x < 3 =⇒ 1 < 3x =⇒ x > 31. So the solution in this case is (0, ∞) ∩ ( 13 , ∞) = ( 31 , ∞). If x < 0, then 1x < 3 =⇒ 1 > 3x =⇒ x < 31. So in this case (−∞, 0) ∩ (−∞, 13 ) = (−∞, 0). Therefore, the solution to the original inequality is (−∞, 0) ∪ ( 13 , ∞). (c)

13. Find y′′ if y =

x2 − 1 . 2 + x2

Solution. y′ = y′′ =

4x + 2x3 − 2x3 + 2x 2x(2 + x2 ) − (x2 − 1)(2x) 6x = = 2 2 (2 + x2 )2 (2 + x ) (2 + x2 )2 6(2 − 3x2 ) 6(2 + x2 )[2 + x2 − 4x2 ] 6(2 + x2 )2 − 6x · 2(2 + x2 )(2x) = = . (2 + x2 )4 (2 + x2 )4 (2 + x2 )3

14. Let g(x) = tan(x + f (x2 + x)), where f is a differentiable function. Given that f (0) = find f ′ (0). Solution. g ′ (x) = [sec2 (x + f (x2 + x))][1 + (f ′ (x2 + x))(2x + 1)]. Letting x = 0, we get g ′ (0) = [sec2 (0 + f (0))][1 + (f ′ (0))(1)] −3 = [sec2 (π/3)][1 + f ′ (0)] = 4(1 + f ′ (0)) =⇒ f ′ (0) = −

3 7 −1=− . 4 4

15. Find all vertical and horizontal asymptotes.

(a) f (x) =

√

5x + 2x2 2x − 7

(b) g(x) =

7ex ex − 3

(c) h(x) = ln(x2 − 2)

π 3

and g ′ (0) = −3,

Page 9 (a) Horizontal asymptotes: q q √ √ √ 5 5x+2x2 2 2 2 x +2 x 5x + 2x 0+2 = lim = lim = lim = 2x−7 x→∞ x→∞ x→∞ 2 − 7 2x − 7 2 − 0 2 qx qx √ √ 5 √ 5x+2x2 2 x +2 x2 5x + 2x2 0+2 = lim . =− = lim lim =− 2x−7 x→−∞ x→−∞ −2 + 7 x→−∞ 2x − 7 2 − 0 2 −x x

Solution.

Therefore, y =

√ √ 2/2 and y = − 2/2.

Vertical asymptotes: Since the bottom = 0 at 7/2, √ 5x + 2x2 =∞ lim+ 2x − 7 x→ 27 since the top →

since the top → Therefore, x =

√ 42 and the bottom → 0 positively, and √ 5x + 2x2 = −∞ lim 2x − 7 x→ 72 − √ 42 and the bottom → 0 negatively,

7 2.

(b) Horizontal asymptotes: 7ex lim x = lim x→∞ x→∞ e − 3 lim

x→−∞

7ex ex ex ex

−

= lim

3 ex

x→∞

7 7 =7 = 1−0 1 − e3x

7ex 0 7·0 = − = 0. = 3 ex − 3 0 − 3

Therefore, y = 7 and y = 0. Vertical asymptotes: We note that the bottom is zero at x = ln 3. Then lim +

x→ln 3

7ex =∞ −3

ex

since the top → 7eln 3 = 7 · 3 = 21 and the bottom → 0 positively, and lim

x→ln 3−

7ex = −∞ −3

ex

since the top → 7eln 3 = 7 · 3 = 21 and the bottom → 0 negatively. Therefore, x = ln 3. (c) Horizontal asymptotes: lim ln(x2 − 2) = ∞

x→∞

and

lim ln(x2 − 2) = ∞

x→−∞

since x2 − 2 → ∞. Therefore, there are no horizontal asymptotes. √ √ Vertical asymptotes: Since the domain is (−∞, − 2) ∪ ( 2, ∞), we calculate the following limits. lim √ − x→− 2

ln(x2 − 2) = −∞ and

lim √ + ln(x x→ 2

√ √ since x2 − 2 → 0 positively. Therefore, x = − 2 and x = 2.

2

− 2) = −∞...