Calculus Midterm Exam

Calculus Midterm Exam

1. (a) (6 points) Find lim

x→π/6

sin(x +¹₃π_{) − 1} x−¹6π

Solution: By setting y= x −¹₆π, lim

x→π/6

sin(x +¹₃π_{) − 1} x−¹₆π ^{= limy→0}

sin(y +¹₂π_{) − 1} y

= lim

y→0

sin y cos¹₂π+ cos y sin¹₂π_{− 1}

y = lim

y→0

cos y− 1

y = lim

y→0

−sin²y y¡ cos y + 1¢

= − lim

y→0

sin y y lim

y→0

sin y cos y+ 1 = 0 (b) (6 points) Does lim

x→0cos¡_{πsin x}

4x ¢ exist ? If so, what is the limit?

Solution: Since cos x is continuous everywhere,

xlim→0cos¡_{πsin x}

4x ¢ = cos ¡ lim

x→0 πsin x

4x ¢ = cos ¡^π₄ lim

x→0 sin x

x ¢ = cos^π₄ = ^√₂² (c) (8 points) Let f be the Dirichlet function

f(x) =

(1, x rational 0, x irrational.

Find lim

x→0x¡5 f (x) + 3¢.

Solution: Since 0≤ |x¡5 f (x) + 3¢| ≤ 8|x| for all x, and lim

x→08|x| = 0, by the Squeeze theorem, we have lim

x→0|x¡5 f (x) + 3¢| = 0, and therefore, lim

x→0x¡5 f (x) + 3¢ = 0.

2. (8 points) Assume that

f(0) = 1, f⁰(0) = 3, f (1) = 0, f⁰(1) = 3, f (2) = 2, f⁰(2) = 2,

g(0) = 1, g⁰(0) = 1, g(1) = 2, g⁰(1) = 2, g(2) = 1, g⁰(2) = 3, h(0) = 1, h⁰(0) = 2, h(1) = 2, h⁰(1) = 2, h(2) = 0, h⁰(2) = 2.

Find( f ◦ h ◦ g)⁰(2).

Solution: ( f ◦ h ◦ g)⁰(2) = f⁰(h ◦ g(2))h⁰(g(2)) g⁰(2) = f⁰(2) h⁰(1) g⁰(2) = 2 · 2 · 3 = 12.

3. (8 points) Find equations for the tangent and normal lines to the curve

2x³+ 2y³= 9xy at the point(1, 2).

Calculus Midterm Exam

Solution: _dx^d¡2x³+ 2y³¢ = _dx^d¡9xy¢

=⇒ 6x²+ 6y^{2 dy}_dx = 9y + 9x^dy_dx holds at each point on the curve.

At the point(1, 2), we get 6 + 24^dy_dx = 18 + 9^dy_dx =⇒^dydx =⁴₅.

Thus, at the point(1, 2), the tangent and normal lines to the curve have equations y− 2 = ⁴5(x − 1), and y − 2 = −⁵4(x − 1), respectively.

4. (a) (6 points) Let f be continuous at c. Prove that if f(c) > 0, then there exists δ > 0 such that f(x) > 0 for all x ∈ (c −δ, c +δ).

Solution: Since f(c) > 0 =⇒ ^f(c)₂ > 0.

By takingε = ^f(c)₂ , theε₋δ definition of the continuity of f at c implies that there exists a δ > 0, such that

if x∈ (c −δ, c +δ) then | f (x) − f (c)| < ^f(c)2

=⇒ −^f(c)₂ < f (x) − f (c)

=⇒ f (x) > ^f(c)2 > 0.

(b) (6 points) Prove that if lim

x→0 f(x)

x exists, then lim

x→0f(x) = 0.

Solution: Since lim

x→0 f(x)

x exists, and lim

x→0x= 0, 0 =¡ lim

x→0 f(x)

x ¢ ¡ lim

x→0x¢ = lim

x→0

¡f(x)

x x¢ = lim

x→0f(x).

5. Let f be continuous on(a, b) and c ∈ (a,b). Suppose that f is differentiable on (a,c) ∪ (c,b).

(a) (8 points) Prove that if lim

x→cf⁰(x) exists, then f⁰(c) exists.

Solution:

c x

a b

Need to prove that lim

x→c

f(x) − f (c)

x− c exists.

Assuming that x∈ (a,c)∪(c,b), since f is continuous on (a,b) and is differentiable on (a,c)∪

(c, b), the Mean Value theorem implies that there exists a zx lying between x and c, such that limx→c

f(x) − f (c) x− c = lim

x→c

f⁰(zx)(x − c) x− c = lim

x→cf⁰(zx) = lim

x→c f⁰(x), where we have used the fact that limx→cz_x= c, since zx lies between x and c.

Remark: The proof above show that if f(x) =

(g(x), x < c

h(x), x > c,

where g and h are differentiable on x< c, and x > c, respectively, then lim

x→c⁻g⁰(x) = lim

x→c⁻

f(x) − f (c)

x− c , and lim

x→c⁺h⁰(x) = lim

x→c⁺

f(x) − f (c)

x− c by the Mean Value theorem.

This implies that if both lim

x→c⁻ f⁰(x) and lim

x→c⁺ f⁰(x) exist, and lim

x→c⁻f⁰(x) 6= lim

x→c⁺ f⁰(x), then f⁰(c) does not exist.

Calculus Midterm Exam

x→cf⁰(x) need not exist, even if f⁰(c) exists.

Solution: Let f(x) =

(x²sin(¹_x), x 6= 0

0, x= 0.

=⇒ lim

x→0

x²sin(¹_x) − 0

x− 0 = 0 by using the Squeeze theorem. Thus, f⁰(0) exists.

Now f⁰(x) =

(2x sin(¹_x) − cos(¹_x), x 6= 0 f⁰(0) = 0, x= 0.

By taking xn= _nπ¹ , we show that lim

x→0f⁰(x) does not exist.

6. (a) (6 points) Estimate f(3.4) given that f (3) = 2 and f⁰(x) =¡x³+ 5¢1/5

.

Solution: f(3.4) = f (3) + f⁰(3) (3.4 − 3) = 2 + 2 · 0.4 = 2.8

(b) (6 points) Suppose that the graph of f is concave upward on the interval(a, b). For any x, x + h ∈ (a, b), show that f (x + h) ≥ f (x) + f⁰(x)h, i.e. the graph of f lies above its tangent lines.

Solution: By the Mean Value theorem, f(x + h) − f (x) − f⁰(x)h =¡ f⁰(z) − f⁰(x)¢h, where z is a point lying between x+ h and h. Since the graph of f is concave upward on the interval (a, b), f⁰is increasing on(a, b). Thus,

f⁰(z) − f⁰(x)

(> 0, if h > 0, i.e. x + h > z > x.

< 0, if h < 0, i.e. x + h < z < x.

Therefore, f(x + h) − f (x) − f⁰(x)h =¡ f⁰(z) − f⁰(x)¢h ≥ 0.

7. (10 points) A 13-foot ladder leans against the side of a building, forming an angleθ with the ground.

Given that the foot of the ladder is being pulled away from the building at the rate of 0.1 feet per second, what is the rate of change ofθ when the top of the ladder is 12 feet above the ground?

Solution:

θ

y

x

dx dt = 0.1

Let y= the distance from top of the ladder to the ground;

x= the distance from the building to the foot of the ladder.

Thus, dx

dt = 0.1, and cosθ = x 13.

Calculus Midterm Exam

=⇒ −sinθ^dθ

dt = ₁₃¹ dx

dt = 0.1 · 1 13. When y= 12, sinθ = 12

13=⇒ dθ

dt = − 1 120.

8. (16 points) Let f(x) = sin 2x + 4 sin x − x, where x ∈£ −π 2,3π

2 ¤.

(a) Find the intervals of increasing or decreasing, local extreme values.

Solution: f⁰(x) = 2 cos 2x + 4 cos x − 1 =¡2 cos x − 1¢¡2cosx + 3), f⁰(x) = 0 ⇔ x = −π

3, π 3.

f⁰(x) > 0 on¡ −^π₃,^π₃¢ and f⁰(x) < 0 on£ −^π₂, −^π3¢ ∪ ¡^π₃,^3π₂¤

f is increasing on¡ −^π₃,^π₃¢ and f is decreasing on £ −^π₂, −^π₃¢ ∪ ¡^π₃,^3π₂ ¤

And, f(−^π3) = −^5√2³+^π₃ is a local minimum, f(^π₃) =^5√₂³−^π3 is a local maximum (b) Find the intervals of concave upward or downward, the points of inflection.

Solution: f⁰⁰(x) = −4sin2x − 4sinx = −4sinx¡2 cos x + 1¢

f⁰⁰(x) = 0 ⇔ x = 0, π, 2π 3 , 4π

3

f⁰⁰(x) > 0 on¡ −^π₂, 0¢ ∪ ¡²₃^π,π¢ ∪ ¡⁴₃^π,³₂^π¢ and f⁰⁰(x) < 0 on¡0,²₃^π¢ ∪ ¡π,⁴₃^π¢

The graph of f is concave upward on¡ −^π₂, 0¢ ∪ ¡²₃^π,π¢ ∪ ¡⁴₃^π,³₂^π¢ and is concave downward on¡0,²₃^π¢ ∪ ¡π,⁴₃^π¢

The inflection points are(0, 0), (π^,π^{), ¡2π}₃ ^,3√₂³−^2π₃ ¢, ¡^4π₃ , −^3√₂³−^4π₃ ¢ (c) Sketch the graph of f.

Solution:

x y

−^π₃ 0 π

3 2π

3 π 4π

−^π₂ 3 ^3π₂