Unit 2: Part B Solutions (Matrices and Calculus)

Question 1

(i) Sketch the graph of the function f(x) = 1+x,     x < -1 f(x) = x2,      -1 ≤ x ≤ 1 f(x) = 2-x,     x ≥ 1 and use it to determine the values of "a" for which limx→a f(x).

(ii) Differentiate y = (2x+1)5(x3-x+1)4.

Solution

Part (i): Graph and Limits
1. Graph Sketch:

  • For x < -1, the graph is the line y = 1+x. This is a line ending at the point (-1, 0) (an open circle).
  • From x = -1 to x = 1, the graph is the parabola y = x2. It starts at (-1, 1) (a closed circle), passes through (0, 0), and ends at (1, 1) (a closed circle).
  • For x ≥ 1, the graph is the line y = 2-x. This line starts at (1, 1) (at the same point the parabola ended) and goes down with a slope of -1.
2. Limits: We only need to check the points where the function definition changes: a = -1 and a = 1.
  • At a = -1:
    Left-hand limit: limx→-1- f(x) = limx→-1- (1+x) = 0
    Right-hand limit: limx→-1+ f(x) = limx→-1+ (x2) = 1
    Since 0 ≠ 1, the limit does not exist at a = -1.
  • At a = 1:
    Left-hand limit: limx→1- f(x) = limx→1- (x2) = 1
    Right-hand limit: limx→1+ f(x) = limx→1+ (2-x) = 1
    Since the left and right limits are equal, the limit exists at a = 1.

(i) The limit limx→a f(x) exists for all real numbers 'a' except a = -1.

Part (ii): Differentiation
We use the Product Rule: d/dx(uv) = u'v + uv'.

  • Let u = (2x+1)5. Then u' = 5(2x+1)4 · 2 = 10(2x+1)4.
  • Let v = (x3-x+1)4. Then v' = 4(x3-x+1)3 · (3x2-1).
dy/dx = [10(2x+1)4] · (x3-x+1)4 + (2x+1)5 · [4(x3-x+1)3(3x2-1)]
Factor out the common terms: 2(2x+1)4 and (x3-x+1)3.
dy/dx = 2(2x+1)4(x3-x+1)3 [ 5(x3-x+1) + (2x+1) · 2(3x2-1) ]
Simplify the expression in the brackets:
[ 5x3 - 5x + 5 + (4x+2)(3x2-1) ]
[ 5x3 - 5x + 5 + (12x3 - 4x + 6x2 - 2) ]
[ 17x3 + 6x2 - 9x + 3 ]

(ii) dy/dx = 2(2x+1)4(x3-x+1)3(17x3 + 6x2 - 9x + 3)

Question 2

(i) If x2 + y2 = 25, then find dy/dx and also find an equation of the tangent line to the curve at the point (3, 4).

(ii) If f(x) = xex then find f'(x). Also find the n-th derivative fn(x).

Solution

Part (i): Implicit Differentiation and Tangent Line
1. Find dy/dx: Differentiate x2 + y2 = 25 with respect to x.
2x + 2y · (dy/dx) = 0
2y (dy/dx) = -2x
dy/dx = -2x / 2y = -x/y
2. Find Tangent Line at (3, 4):
First, find the slope (m) at (3, 4):
m = dy/dx |(3,4) = -3 / 4
Using the point-slope form y - y1 = m(x - x1):
y - 4 = (-3/4)(x - 3)
Multiply by 4: 4(y - 4) = -3(x - 3)
4y - 16 = -3x + 9
3x + 4y = 25

(i) dy/dx = -x/y. The tangent line at (3, 4) is 3x + 4y = 25.

Part (ii): n-th Derivative
We find the first few derivatives to identify a pattern.

  • f(x) = xex
  • f'(x) = (1)ex + (x)ex = ex(1 + x)
  • f''(x) = (ex)(1+x) + (ex)(1) = ex(1 + x + 1) = ex(2 + x)
  • f'''(x) = (ex)(2+x) + (ex)(1) = ex(2 + x + 1) = ex(3 + x)
The pattern is clear: the n-th derivative adds n to x.

(ii) f'(x) = ex(1+x). The n-th derivative is fn(x) = ex(n + x).

Question 3

(i) For what values of a and b, is f(x) = ax-b,               x ≤ -1 f(x) = 4x3-12x2+1,   -1 < x < 1 f(x) = 3,                   x ≥ 1 continuous at every x?

(ii) Find dy/dx, if y = (sec x) / (1 + tan x).

Solution

Part (i): Continuity
For f(x) to be continuous everywhere, it must be continuous at the "break" points x = -1 and x = 1.
1. Continuity at x = -1:
limx→-1- f(x) = limx→-1- (ax - b) = a(-1) - b = -a - b
limx→-1+ f(x) = limx→-1+ (4x3 - 12x2 + 1) = 4(-1)3 - 12(-1)2 + 1 = -4 - 12 + 1 = -15
For continuity, we must have: -a - b = -15 or a + b = 15.
2. Continuity at x = 1:
limx→1- f(x) = limx→1- (4x3 - 12x2 + 1) = 4(1)3 - 12(1)2 + 1 = 4 - 12 + 1 = -7
limx→1+ f(x) = limx→1+ (3) = 3
The left-hand limit (-7) does not equal the right-hand limit (3).

(i) Since the limit at x = 1 does not exist (as -7 ≠ 3), it is impossible for the function to be continuous at x = 1. Therefore, there are no values of a and b that can make this function continuous at every x.

Part (ii): Differentiation
We use the Quotient Rule: d/dx(u/v) = (u'v - uv') / v2.

  • Let u = sec x. Then u' = sec x tan x.
  • Let v = 1 + tan x. Then v' = sec2 x.
dy/dx = [ (sec x tan x)(1 + tan x) - (sec x)(sec2 x) ] / (1 + tan x)2
Expand the numerator:
[ sec x tan x + sec x tan2 x - sec3 x ] / (1 + tan x)2
Factor out sec x from the numerator:
[ sec x (tan x + tan2 x - sec2 x) ] / (1 + tan x)2
Use the identity 1 + tan2 x = sec2 x, which means tan2 x - sec2 x = -1.
[ sec x (tan x - 1) ] / (1 + tan x)2

(ii) dy/dx = [ sec x (tan x - 1) ] / (1 + tan x)2

Question 4

(i) Find the absolute maximum and absolute minimum value of the function f(x) = 3x4, [-2, 3].

(ii) Evaluate (1) d/dx(3x5 log x) and (2) d/dx(x3 / (3x-2)).

(iii) Find dy/dx, if y = (sin x)cos x.

Solution

Part (i): Absolute Extrema
1. Find critical points: f'(x) = 12x3.
Set f'(x) = 0 → 12x3 = 0 → x = 0.
2. Test endpoints and critical points: We test the values x = -2, x = 3 (endpoints) and x = 0 (critical point) in the original function f(x) = 3x4.

  • f(-2) = 3(-2)4 = 3(16) = 48
  • f(0) = 3(0)4 = 0
  • f(3) = 3(3)4 = 3(81) = 243

(i) The Absolute Maximum is 243 (at x = 3). The Absolute Minimum is 0 (at x = 0).

Part (ii-1): Derivative (Product Rule)
d/dx(3x5 ln x)   (Assuming log x is natural log, ln x)
= [d/dx(3x5)] · (ln x) + (3x5) · [d/dx(ln x)]
= (15x4)(ln x) + (3x5)(1/x)
= 15x4 ln x + 3x4 = 3x4(5 ln x + 1)

Part (ii-2): Derivative (Quotient Rule)
d/dx( x3 / (3x-2) )
= [ (3x2)(3x-2) - (x3)(3) ] / (3x-2)2
= [ 9x3 - 6x2 - 3x3 ] / (3x-2)2
= [ 6x3 - 6x2 ] / (3x-2)2 = 6x2(x - 1) / (3x - 2)2

Part (iii): Logarithmic Differentiation
Given y = (sin x)cos x.
1. Take the natural log of both sides:
ln y = ln( (sin x)cos x )
ln y = (cos x) · ln(sin x)
2. Differentiate implicitly w.r.t. x (using Product Rule on the right):
(1/y) · (dy/dx) = [d/dx(cos x)] · ln(sin x) + (cos x) · [d/dx(ln(sin x))]
(1/y)(dy/dx) = (-sin x) ln(sin x) + (cos x) · (1/sin x) · (cos x)
(1/y)(dy/dx) = -sin x ln(sin x) + (cos2 x / sin x)
3. Solve for dy/dx by multiplying by y:
dy/dx = y [ -sin x ln(sin x) + (cos2 x / sin x) ]
dy/dx = (sin x)cos x [ -sin x ln(sin x) + (cos2 x / sin x) ]

Question 5

(i) Find critical value of the function f(x) = (x-1) / (x2-x+1).

(ii) Find dy/dx, if y = (sec x) / (1 + tan x).

Solution

Part (i): Critical Values
Critical values occur where f'(x) = 0 or f'(x) is undefined.
1. Find f'(x) using the Quotient Rule:
f'(x) = [ (1)(x2-x+1) - (x-1)(2x-1) ] / (x2-x+1)2
Expand the numerator:
f'(x) = [ (x2-x+1) - (2x2 - 3x + 1) ] / (x2-x+1)2
f'(x) = [ x2 - x + 1 - 2x2 + 3x - 1 ] / (x2-x+1)2
f'(x) = [ -x2 + 2x ] / (x2-x+1)2
2. Find where f'(x) = 0:
-x2 + 2x = 0 → -x(x - 2) = 0
This gives x = 0 and x = 2.
3. Find where f'(x) is undefined:
This occurs if the denominator is zero. (x2-x+1) = 0.
Using the discriminant (b2-4ac): (-1)2 - 4(1)(1) = 1 - 4 = -3. Since it's negative, the denominator has no real roots and is never zero.

(i) The critical values are x = 0 and x = 2.

Part (ii): Differentiation

Note: This is identical to Question 3(ii).

We use the Quotient Rule: d/dx(u/v) = (u'v - uv') / v2.
  • u = sec x,    u' = sec x tan x
  • v = 1 + tan x,   v' = sec2 x
dy/dx = [ (sec x tan x)(1 + tan x) - (sec x)(sec2 x) ] / (1 + tan x)2
Factor out sec x in the numerator:
dy/dx = [ sec x (tan x(1 + tan x) - sec2 x) ] / (1 + tan x)2
dy/dx = [ sec x (tan x + tan2 x - sec2 x) ] / (1 + tan x)2
Using the identity tan2 x - sec2 x = -1:
dy/dx = [ sec x (tan x - 1) ] / (1 + tan x)2

(ii) dy/dx = [ sec x (tan x - 1) ] / (1 + tan x)2

Question 6

For f(x) = 2x3 + 3x2 - 36x, find:
(i) Critical points.
(ii) On what interval is f increasing or decreasing?
(iii) At what points if any, does f assume local maximum and minimum values?
(iv) Find intervals of concavity and the inflection points.

Solution

f(x) = 2x3 + 3x2 - 36x
f'(x) = 6x2 + 6x - 36 = 6(x2 + x - 6) = 6(x+3)(x-2)
f''(x) = 12x + 6 = 6(2x+1)

(i) Critical points:
Set f'(x) = 0 → 6(x+3)(x-2) = 0.
The critical points are x = -3 and x = 2.

(ii) Increasing/Decreasing: We test intervals around the critical points.

  • Interval (-∞, -3): Let x = -4. f'(-4) = 6(-1)(-6) = + (Increasing)
  • Interval (-3, 2): Let x = 0. f'(0) = 6(3)(-2) = - (Decreasing)
  • Interval (2, ∞): Let x = 3. f'(3) = 6(6)(1) = + (Increasing)

(iii) Local Max/Min:

  • At x = -3, f(x) changes from Increasing to Decreasing → Local Maximum.
    f(-3) = 2(-27) + 3(9) - 36(-3) = -54 + 27 + 108 = 81.
  • At x = 2, f(x) changes from Decreasing to Increasing → Local Minimum.
    f(2) = 2(8) + 3(4) - 36(2) = 16 + 12 - 72 = -44.

(iv) Concavity/Inflection Points:
Set f''(x) = 0 → 12x + 6 = 0 → x = -1/2.

  • Interval (-∞, -1/2): Let x = -1. f''(-1) = 12(-1) + 6 = - (Concave Down)
  • Interval (-1/2, ∞): Let x = 0. f''(0) = 12(0) + 6 = + (Concave Up)
The concavity changes at x = -1/2, so there is an inflection point.
f(-1/2) = 2(-1/8) + 3(1/4) - 36(-1/2) = -1/4 + 3/4 + 18 = 1/2 + 18 = 18.5

(i) Critical Points: x = -3, x = 2
(ii) Increasing: (-∞, -3) U (2, ∞). Decreasing: (-3, 2).
(iii) Local Max: 81 at x = -3. Local Min: -44 at x = 2.
(iv) Concave Up: (-1/2, ∞). Concave Down: (-∞, -1/2). Inflection Point: (-0.5, 18.5).

Question 7

For f(x) = 2 + 2x2 - x4, find:
(i) Critical points.
(ii) On what interval is f increasing or decreasing?
(iii) At what points if any, does f assume local maximum and minimum values?
(iv) Find the intervals of concavity and the inflection points.

Solution

f(x) = 2 + 2x2 - x4
f'(x) = 4x - 4x3 = 4x(1 - x2) = 4x(1-x)(1+x)
f''(x) = 4 - 12x2 = 4(1 - 3x2)

(i) Critical points:
Set f'(x) = 0 → 4x(1-x)(1+x) = 0.
The critical points are x = 0, x = 1, and x = -1.

(ii) Increasing/Decreasing:

  • Interval (-∞, -1): Let x = -2. f'(-2) = 4(-2)(1-4) = + (Increasing)
  • Interval (-1, 0): Let x = -0.5. f'(-0.5) = 4(-0.5)(1-0.25) = - (Decreasing)
  • Interval (0, 1): Let x = 0.5. f'(0.5) = 4(0.5)(1-0.25) = + (Increasing)
  • Interval (1, ∞): Let x = 2. f'(2) = 4(2)(1-4) = - (Decreasing)

(iii) Local Max/Min:

  • At x = -1, (Inc → Dec) → Local Maximum.
    f(-1) = 2 + 2(-1)2 - (-1)4 = 2 + 2 - 1 = 3.
  • At x = 0, (Dec → Inc) → Local Minimum.
    f(0) = 2 + 0 - 0 = 2.
  • At x = 1, (Inc → Dec) → Local Maximum.
    f(1) = 2 + 2(1)2 - (1)4 = 2 + 2 - 1 = 3.

(iv) Concavity/Inflection Points:
Set f''(x) = 0 → 4(1 - 3x2) = 0 → x2 = 1/3 → x = ±1/√3.

  • Interval (-∞, -1/√3): Let x = -1. f''(-1) = 4 - 12(1) = - (Concave Down)
  • Interval (-1/√3, 1/√3): Let x = 0. f''(0) = 4 - 0 = + (Concave Up)
  • Interval (1/√3, ∞): Let x = 1. f''(1) = 4 - 12(1) = - (Concave Down)
The concavity changes at x = ±1/√3.
f(±1/√3) = 2 + 2(1/3) - (1/3)2 = 2 + 2/3 - 1/9 = 18/9 + 6/9 - 1/9 = 23/9.

(i) Critical Points: x = -1, 0, 1
(ii) Increasing: (-∞, -1) U (0, 1). Decreasing: (-1, 0) U (1, ∞).
(iii) Local Max: 3 at x = -1 and x = 1. Local Min: 2 at x = 0.
(iv) Concave Up: (-1/√3, 1/√3). Concave Down: (-∞, -1/√3) U (1/√3, ∞).
Inflection Points: (-1/√3, 23/9) and (1/√3, 23/9).

Question 8

(i) Find the values of a and b that make f continuous on (-∞, ∞) if f(x) = (xi-bi)/(x-2),   if x < 2    (Assuming typo for (x2-4)/(x-2)) f(x) = ax2-bx+3,     if 2 ≤ x ≤ 3 f(x) = 2x-a+b,          if x ≥ 3

(ii) Find dy/dx if y = x2e2x(x2+1)4.

Solution

Part (i): Continuity

Note: The term (xi-bi)/(x-2) is assumed to be the common limit problem (x2-4)/(x-2), which means f(x) = x+2 for x < 2.

1. Continuity at x = 2:
limx→2- f(x) = limx→2- (x2-4)/(x-2) = limx→2- (x+2) = 4
limx→2+ f(x) = f(2) = a(2)2 - b(2) + 3 = 4a - 2b + 3
Equation 1: 4a - 2b + 3 = 4  →  4a - 2b = 1
2. Continuity at x = 3:
limx→3- f(x) = f(3) = a(3)2 - b(3) + 3 = 9a - 3b + 3
limx→3+ f(x) = 2(3) - a + b = 6 - a + b
Equation 2: 9a - 3b + 3 = 6 - a + b  →  10a - 4b = 3
3. Solve the system of equations:
(1) 4a - 2b = 1
(2) 10a - 4b = 3
From (1), multiply by 2: 8a - 4b = 2.
Subtract this new equation from (2):
(10a - 4b) - (8a - 4b) = 3 - 2
2a = 1  →  a = 1/2
Substitute a = 1/2 into (1): 4(1/2) - 2b = 1 → 2 - 2b = 1 → 1 = 2b  →  b = 1/2

(i) a = 1/2 and b = 1/2.

Part (ii): Logarithmic Differentiation
Given y = x2e2x(x2+1)4
1. Take the natural log of both sides:
ln y = ln(x2) + ln(e2x) + ln((x2+1)4)
ln y = 2 ln x + 2x + 4 ln(x2+1)
2. Differentiate implicitly w.r.t. x:
(1/y)(dy/dx) = 2/x + 2 + 4 · (1 / (x2+1)) · (2x)
(1/y)(dy/dx) = 2/x + 2 + (8x / (x2+1))
3. Solve for dy/dx by multiplying by y:

(ii) dy/dx = x2e2x(x2+1)4 [ 2/x + 2 + 8x/(x2+1) ]

Question 9

For f(x) = x4 - 2x3 + 3, find:
(i) Critical points.
(ii) On what interval is f increasing or decreasing?
(iii) At what points if any, does f assume local maximum and minimum values?
(iv) Find the intervals of concavity and the inflection points.

Solution

f(x) = x4 - 2x3 + 3
f'(x) = 4x3 - 6x2 = 2x2(2x - 3)
f''(x) = 12x2 - 12x = 12x(x - 1)

(i) Critical points:
Set f'(x) = 0 → 2x2(2x - 3) = 0.
The critical points are x = 0 and x = 3/2.

(ii) Increasing/Decreasing:

  • Interval (-∞, 0): Let x = -1. f'(-1) = 2(1)(-5) = - (Decreasing)
  • Interval (0, 3/2): Let x = 1. f'(1) = 2(1)(-1) = - (Decreasing)
  • Interval (3/2, ∞): Let x = 2. f'(2) = 2(4)(1) = + (Increasing)

(iii) Local Max/Min:

  • At x = 0, (Dec → Dec) → Neither a max nor a min (it is a saddle point).
  • At x = 3/2, (Dec → Inc) → Local Minimum.
    f(3/2) = (3/2)4 - 2(3/2)3 + 3 = 81/16 - 2(27/8) + 3 = 81/16 - 108/16 + 48/16 = 21/16.

(iv) Concavity/Inflection Points:
Set f''(x) = 0 → 12x(x - 1) = 0 → x = 0 and x = 1.

  • Interval (-∞, 0): Let x = -1. f''(-1) = 12(-1)(-2) = + (Concave Up)
  • Interval (0, 1): Let x = 0.5. f''(0.5) = 12(0.5)(-0.5) = - (Concave Down)
  • Interval (1, ∞): Let x = 2. f''(2) = 12(2)(1) = + (Concave Up)
Concavity changes at x = 0 and x = 1.
f(0) = 3
f(1) = 14 - 2(1)3 + 3 = 1 - 2 + 3 = 2

(i) Critical Points: x = 0, x = 3/2
(ii) Increasing: (3/2, ∞). Decreasing: (-∞, 0) U (0, 3/2).
(iii) Local Min: 21/16 at x = 3/2. No Local Max.
(iv) Concave Up: (-∞, 0) U (1, ∞). Concave Down: (0, 1).
Inflection Points: (0, 3) and (1, 2).

Question 10

For f(x) = sin x + cos x, [0, 2Π], find:
(i) Critical points.
(ii) On what interval is f increasing or decreasing?
(iii) At what points if any, does f assume local maximum and minimum values?
(iv) Find the intervals of concavity and the inflection points.

Solution

f(x) = sin x + cos x
f'(x) = cos x - sin x
f''(x) = -sin x - cos x

(i) Critical points:
Set f'(x) = 0 → cos x - sin x = 0 → cos x = sin x → tan x = 1.
In the interval [0, 2Π], this occurs at x = Π/4 and x = 5Π/4.

(ii) Increasing/Decreasing:

  • Interval [0, Π/4): Let x = 0. f'(0) = cos(0) - sin(0) = 1 = + (Increasing)
  • Interval (Π/4, 5Π/4): Let x = Π. f'(Π) = cos(Π) - sin(Π) = -1 = - (Decreasing)
  • Interval (5Π/4, 2Π]: Let x = 2Π. f'(2Π) = cos(2Π) - sin(2Π) = 1 = + (Increasing)

(iii) Local Max/Min:

  • At x = Π/4, (Inc → Dec) → Local Maximum.
    f(Π/4) = sin(Π/4) + cos(Π/4) = (1/√2) + (1/√2) = 2/√2 = √2.
  • At x = 5Π/4, (Dec → Inc) → Local Minimum.
    f(5Π/4) = sin(5Π/4) + cos(5Π/4) = (-1/√2) + (-1/√2) = -2/√2 = -√2.

(iv) Concavity/Inflection Points:
Set f''(x) = 0 → -sin x - cos x = 0 → tan x = -1.
In the interval [0, 2Π], this occurs at x = 3Π/4 and x = 7Π/4.

  • Interval [0, 3Π/4): Let x = Π/2. f''(Π/2) = -sin(Π/2) - cos(Π/2) = -1 = - (Concave Down)
  • Interval (3Π/4, 7Π/4): Let x = Π. f''(Π) = -sin(Π) - cos(Π) = 1 = + (Concave Up)
  • Interval (7Π/4, 2Π]: Let x = 2Π. f''(2Π) = -sin(2Π) - cos(2Π) = -1 = - (Concave Down)
Inflection points are at x = 3Π/4 and x = 7Π/4.
f(3Π/4) = sin(3Π/4) + cos(3Π/4) = (1/√2) + (-1/√2) = 0
f(7Π/4) = sin(7Π/4) + cos(7Π/4) = (-1/√2) + (1/√2) = 0

(i) Critical Points: x = Π/4, x = 5Π/4
(ii) Increasing: [0, Π/4) U (5Π/4, 2Π]. Decreasing: (Π/4, 5Π/4).
(iii) Local Max: √2 at x = Π/4. Local Min: -√2 at x = 5Π/4.
(iv) Concave Up: (3Π/4, 7Π/4). Concave Down: [0, 3Π/4) U (7Π/4, 2Π].
Inflection Points: (3Π/4, 0) and (7Π/4, 0).