Solutions For Calculus Midterm

Solutions For Calculus Midterm

1. (a)

f (x) = x 3^x

f (x) = 3^x(x)⁰− x(3^x)⁰

(3^x)² = 3^x− x3^xln 3

3^2x = 1− x ln 3 3^x (b)

f⁰(x) = 2(ln x)(ln x)⁰ = 2 ln x x (c)

f⁰(x) = ln(3x⁴ + 5)(e^x2)⁰− e^x2[ln(3x⁴+ 5)]⁰

[ln(3x⁴+ 5)]² = 2xe^x2ln(3x⁴+ 5)− ^12x_3x⁴³₊₅^ex2 [ln(3x⁴+ 5)]² 2. (a) Applying L’Hospital’s rule twice we have,

xlim→∞

(ln x)²

x² = lim

x→∞

2·_x¹ln x

2x = lim

x→∞

ln x

x² = lim

x→∞

1 x

2x = 0 (b) If ex²sin(1

x) converges as x→ 0, then any subsequences of ex²sin(1

x) will converge to the limit as x→ 0. In other words, if there are two subsequences of ex²sin(1

x) converge to different limits as x → 0, then it concludes that the limit of ex²sin(1

x) does not exist. Consider two subsequences {e^{(nπ1 )2}sin(nπ)} and {e^{(π2+2nπ1 )2}sin(π

2 + 2nπ)}, then we have that the limit of ex²sin(1

x) does not exist since {e^{(nπ1 )2}sin(nπ)} converges to 0 while {e^{(π2+2nπ1 )2}sin(π

2 + 2nπ)} converges to 1.

x _π¹ _2π¹ _3π¹ _4π¹ → 0

1

x π 2π 3π 4π ...

sin(¹_x) 0 0 0 0 ...

e²sin(¹_x) 0 0 0 0 ...

x _π² _5π² _9π² _13π² → 0

1 x

π 2

5π 2

9π 2

13π

2 ...

sin(¹_x) 1 1 1 1 ...

e²sin(¹_x) e^(π2)2 e^{(5π2 )2} e^{(9π2 )2} e^{(13π2 )2} → 1 (c) Applying L’Hospital’s rule we have,

xlim→∞x(e^1x − 1) = lim

x→∞

e^1x − 1

1 x

= lim

x→∞

−_x¹²e^1x

−_x¹² = lim

x→∞e^1x = 1

3. (a) Let

y = √

5− x² y² = 5− x² x² = 5− y²

x = √

5− y²

∴ f⁻¹(y) = √

5− y², 5− y² ≥ 0 (b)

d

dyf⁻¹(y) = 1

2(5− y²)^−1/2(5− y²)⁰ = −2y 2√

5− y² = √−y 5− y² (c)

f⁰(x) = 1

2(5− x²)^−1/2(5− x²)⁰ = √−x 5− x² d

dxf⁻¹(y) = 1

−(f⁻¹(y))

√5−(f⁻¹(y))²

= −1

−√

5−y² q

5−(√

5−y²)²

= √−y 5− y²

4. (a) We differentiate both sides of the equation ax²+ by² = 1 with respect to x.

We need to remember that y is a function of x.

d

dx(ax²+ by²) = d dx(1) d

dx(ax²) + d

dx(by²) = d

dx(1) (the derivative of a sum is the sum of the derivatives) 2ax + 2bydy

dx = 0 (the power rule and the chain rule ) We find that dy

dx =−2ax

2by =−ax by

Differentiateing both sides of this equation with respect to x, we have d

dx[dy

dx] = d dx[−ax

by] d²y

dx² = −aby + abx^dy_dx (by)² Substituting −^ax_by for ^dy_dx, we obtain

d²y

dx² = −aby + abx(−^ax_by)

(by)² = −aby −^a2_y^x2

(by)² = −aby²− a²x²

b²y³ = −a(by²+ ax²)

b²y³ =− a b²y³ since ax² + by² = 1

(b) We have the largest curvature when y = 0. That is, (x, y) = (±a, 0)

5. (a) If f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c ∈ (a, b) such that ^{f (b)}_b^−f(a)_−a = f⁰(c).

a c c b

)) ( , (a f a

)) ( , (b f b

(b) f is continuous on [−π, π] since sin x and −x are both continuous on [−π, π].

f is differentiable on (−π, π) since sin x and −x are both differentiable on (−π, π).

Then there exists at least one number c ∈ (a, b) such that f⁰(c) = ^{f (−π)−f(π)}_π−(−π) = ⁽_π^−π)−π_−(−π) =

−1, thanks to the M.V.T.

Solving f⁰(c) = cos c− 1 = −1 for c ∈ (−π, π), we find c = ±π/2.

6. (a)

f (x) =

{ x²− 9, if −4 ≤ x ≤ −3 or 3 ≤ x ≤ 5;

9− x², if −3 < x < 3.

f⁰(x) =

{ 2x, if −4 ≤ x ≤ −3 or 3 ≤ x ≤ 5;

−2x, if −3 < x < 3.

f⁰⁰(x) =

{ 2, if −4 ≤ x ≤ −3 or 3 ≤ x ≤ 5;

−2, if −3 < x < 3.

i. f is increasing on [−3, 0] ∪ [3, 5]; f is decreasing on [−4, −3] ∪ [0, 3];

f is concave up on [−4, −3] ∪ [3, 5]; f is concave down on [−3, 3];

ii. We find that f⁰(0) = 0 and f⁰⁰(0) =−2 < 0.

Hence, f (x) has a local maximum at x = 0, namely (0, 9).

iii. As the second derivative is not equal to zero for any x, we do not have any inflection points. And f (x) does not have any asymptotes.

iv.

−4 −3 −2 −1 0 1 2 3 4 5

0 2 4 6 8 10 12 14 16

(b)

f⁰(x) = −xe^−x22 , x∈ R

f⁰⁰(x) = (−1)e^−x22 + (−x)(−x)e^−x22

= e^−x22 (x²− 1), x ∈ R

i. f is increasing on (−∞, 0); f is decreasing on (0, +∞); f is concave up on (−∞, −1)∪

(1, +∞); f is concave down on (−1, 1); f⁰(x) = 0 for x = 0, and f⁰⁰(x) = 0 for x =±1.

ii. We find that f⁰(0) = 0 and f⁰⁰(0) < 0.

Hence, f (x) has a local maximum at x = 0, namely (0, 1).

iii. f⁰⁰(x) = 0 for x = ±1. There are two inflection points, one at x = −1, namely (−1, e^−1/2), and the other at x = 1, namely (1, e^−1/2). Moreover, lim_x→−∞f (x) = 0 and limx→+∞f (x) = 0. This shows that y = 0 is a horizontal asymptote.

iv.

-4 -3 -2 -1 0 1 2 3 4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

7. By the definition of g(x), g(x) is continuous for x 6= 2, limx→2⁻g(x) = |4c| + 1, and lim_x→2⁺g(x) = 4− 2c. Therefore,

g is continuous at x = 2

⇔ lim

x→2⁻g(x) = lim

x→2⁺g(x) = g(2)

⇔ |4c| + 1 = 4 − 2c

⇔ c = 1/2, −3/2

8. (a) The slop of the straight line connected by two points (0,1/4) and (90,1) is 1− 1/4

90− 0 = 1 120 The difference equation will be

Nt

N_t+1 − 1/4 = 1

120(N_t− 0) N_t

N_t+1 = N_t+ 30 120

Solve

N = 120N

N + 30 N²− 90N = 0 N (N − 90) = 0

N = 0, 90

Then we find all fixed point 0, 90. To determine stability of an equilibrium, will return to the equilibrium. we will start at a value that is different from the equilibrium and check whether the solution The figure following shows that the fixed point 90 is attractive. Moreover, the trivial fixed point 0 is unstable.

N_t 30 60 80 87.28 89.3 89.82 89.95 N_t+1 60 80 87.28 89.3 89.82 89.95 89.99

0 10 20 30 40 50 60 70 80 90

0 10 20 30 40 50 60 70 80 90

N_t Nt+1