Calculus Midterm Exam

Calculus Midterm Exam

1. Let f (x) =





√

x²+ 9 − 3

x² if x < 0 x sin¹_x if x ≥ 0 (a) (5 points) Find lim

x→0⁻f (x).

Solution: lim

x→0⁻ f (x) = lim

x→0⁻

√x²+ 9 − 3

x² = lim

x→0⁻

x² x²(√

x²+ 9 + 3) = 1 6. (b) (5 points) Find lim

x→0⁺f (x).

Solution: Since 0 ≤ |x sin1

x| ≤ |x|, and lim

x→0⁺|x| = 0 = lim

x→0⁺0, Sandwich theorem implies that lim

x→0⁺|x sin1

x| = 0. Hence lim

x→0⁺x sin1

x = 0, and lim

x→0⁺f (x) = 0.

(c) (5 points) Determine where f (x) is continuous.

Solution: Since p

x²+ 9 − 3, x, sin x are everywhere continuous, and 1

x is continuous for any x 6= 0, f (x) is continuous everywhere except possibly at x = 0.

Now lim

x→0⁻ f (x) 6= lim

x→0⁺ f (x), f (x) is not continuous at x = 0.

Hence, f (x) is continuous on (−∞, 0) ∪ (0,∞).

2. For each x ∈ R, let f (x) = x g(x), where g is defined by

g(x) =

(−1 if x ≤ 0

1 if x > 0

(a) (5 points) Find lim

x→0f (x) (if it exists). Justify your answer.

Solution: Since 0 ≤ | f (x)| = |x||g(x)| = |x|, and lim

x→0|x| = 0 = lim

x→00, Sandwich theorem implies that limx→0| f (x)| = 0. Hence, limx→0f (x) = 0.

(b) (5 points) Show that f (x) is continuous at x = 0.

Solution: By the result of part (a), we have lim_x→0f (x) = 0 = f (0). Therefore, f (x) is contin- uous at x = 0.

(c) (5 points) Determine whether lim

x→0

f (x) − f (0)

x − 0 exists. Justify your answer.

Solution: Since lim

x→0

f (x) − f (0) x − 0 = lim

x→0

xg(x) x = lim

x→0g(x), and lim

x→0⁻g(x) = −1 6= 1 = lim

x→0⁺g(x),

x→0limg(x) does not exist.

(d) (5 points) Show that f (x) is not differentiable at x = 0.

Calculus Midterm Exam November 13, 2006

Solution: By the result of part (c), lim

x→0

f (x) − f (0)

x − 0 does not exist, f (x) is not differentiable at x = 0.

3. Suppose that f⁰(x) = 1 x. (a) (5 points) Find d

dxf (sin²x +√

x + 1 + 3).

Solution: d

dxf (sin²x +√

x + 1 + 3) = f⁰(sin²x +√

x + 1 + 3) d dx

¡sin²x +√

x + 1 + 3¢

=

2 sin x cos x +1

2(x + 1)⁻¹² sin²x +√

x + 1 + 3 .

(b) (5 points) Find d²

dx²f (sin²x +√

x + 1 + 3).

Solution: d

dxf (sin²x +√

x + 1 + 3) = f⁰⁰(sin²x +√

x + 1 + 3)

³d

dx(sin²x +√

x + 1 + 3)

´₂ + f⁰(sin²x +√

x + 1 + 3) d² dx²

¡sin²x +√

x + 1 + 3¢

= −¡

2 sin x cos x +1

2(x + 1)⁻¹² ¢₂

¡sin²x +√

x + 1 + 3¢₂ +2¡

cos²x − sin²x¢

−1

4(x + 1)⁻³² sin²x +√

x + 1 + 3 .

4. Consider the curve x²y − y²x = sin x.

(a) (8 points) Find dy

dx at the point (π^{, 0).}

Solution: Differentiating both sides with respect to x, and using implicit differentiation, we get 2xy + x²dy

dx− 2ydy

dxx − y²= cos x. At (π, 0), we haveπ^2dy

dx= −1, or dy

dx = − 1 π^2. (b) (6 points) Find an equation for the tangent line to the curve at (π^{, 0).}

Solution: By part (a), the tangent line to the curve at (π, 0) has an equation y = − 1

π^{2(x −}π^).

(c) (6 points) Find d²y

dx² at the point (π^{, 0).}

Solution: Differentiating both sides of 2xy + x²dy

dx− 2ydy

dxx − y²= cos x with respect to x, we get 2y + 4xdy

dx+ x²d²y

dx²− 2¡dy dx

¢₂

x − 2yd²y

dx²x − 4ydy

dx = − sin x. At (π, 0), we have 4π^¡−1π²

¢+ π^2d2y

dx²− 2¡−1 π²

¢₂

π= 0. Hence, d²y dx² = 4

π³⁺ 2 π^5.

cont’d on the back

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Calculus Midterm Exam November 13, 2006

5. (a) (5 points) Let f (x) = 3x³− 4x²− x + 2. Use the intermediate value theorem to conclude that f (x) = 0 has a solution in (−1, 0).

Solution: Since f (x) is continuous on [−1, 0], and f (−1) = −4 < 0, f (0) = 2 > 0, the interme- diate value theorem implies that there is a c ∈ (−1, 0) such that f (c) = 0.

(b) (5 points) Suppose that f is differentiable for x ∈ R and, furthermore, that f satisfies f (0) = 0 and f⁰(x) ≤ 2 for all x > 0. Use the mean value theorem to show that f (x) ≤ 2x for all x ≥ 0.

Solution: For any x ≥ 0, the mean value theorem implies that there exists a 0 < c < x such that f (x) = f (x) − f (0) = f⁰(c)(x − 0) = f⁰(c)x ≤ 2x.

6. Let f (x) =2x²− 6

x + 2 for x 6= 0.

(a) (2 points) Show that x = −2 is a vertical asymptote of the curve y = f (x).

Solution: Since lim

x→−2⁻f (x) = lim

x→−2⁻

2x²− 6

x + 2 = −∞, or lim

x→−2⁺ f (x) = lim

x→−2⁺

2x²− 6 x + 2 =∞, x = −2 is a vertical asymptote of the curve y = f (x).

(b) (2 points) Show that y = 2x − 4 is an oblique asymptote of the curve y = f (x).

Solution: Since f (x) = 2x²− 6 x + 2 = 2¡

x²− 4¢ + 2

x + 2 = 2x − 4 + 2

x + 2, and

x→lim∞

¡f (x) − (2x − 4)¢

= lim

x→∞

2

x + 2 = 0⁺, or lim

x→−∞

¡f (x) − (2x − 4)¢

= lim

x→−∞

2

x + 2 = 0⁻, y = 2x − 4 is an oblique asymptote of the curve y = f (x).

(c) (3 points) Find the x-intercepts, i.e. zeros of f (x), and the y-intercept of the graph of y = f (x).

Solution: Since f (x) = 2x²− 6 x + 2 = 2¡

x +√ 3¢¡

x −√ 3¢

x + 2 , x-intercepts are −√ 3,√

3.

The y-intercept is at y = f (0) = −3.

(d) (6 points) Determine where f (x) is increasing and where it is decreasing. Does f (x) have local extreme values?

Solution: By part (c), f (x) = ^2x_x+2²⁻⁶ = 2x − 4 +_x+2² , f⁰(x) = 2 −_(x+2)² 2 =²

¡

(x+2)²−1²¢

(x+2)² = 2(x+3)(x+1) (x+2)² .

We have f⁰(x) > 0 on (−∞, −3) ∪ (−1,∞), and f⁰(x) < 0 on (−3, −2) ∪ (−2, −1).

Hence, f (x) is increasing on (−∞, −3)∪(−1,∞), and f (x) is decreasing on (−3, −2)∪(−2, −1).

At x = −3, since f⁰(−3⁻) > 0, f⁰(−3⁺) < 0, f (−3) = −12 is a local maximum value.

At x = −1, since f⁰(−1⁻) < 0, f⁰(−3⁺) > 0, f (−3) = −12 is a local minimum value.

(e) (6 points) Determine where f (x) is concave up and where it is concave down. Does f (x) have inflection points?

Solution: By part (d), f⁰(x) = 2 −_(x+2)² 2, we have f⁰⁰(x) =_(x+2)⁴ 3. We have f⁰⁰(x) < 0 on (−∞, −2), and f⁰⁰(x) > 0 on (−2,∞).

Hence, f (x) is concave up on (−2,∞), and f (x) is concave down on (−∞, −2).

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Calculus Midterm Exam November 13, 2006 (f) (6 points) Sketch the graph of f (x) together with its asymptotes and inflection points (if they exist).

100

0 50

0 -20

-50

-100 -40

x

20 40

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