Unit 2: Part B Solutions (Matrices and Calculus)

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 \( y = x^2 \). It starts at \((-1, 1)\) (a closed circle), passes through \((0, 0)\), and ends at \((1, 1)\) (a closed circle).
• For \( x \ge 1 \), 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: \( \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (1+x) = 0 \)
  Right-hand limit: \( \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (x^2) = 1 \)
  Since \( 0 \neq 1 \), the limit does not exist at \( a = -1 \).
• At \( a = 1 \):
  Left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2) = 1 \)
  Right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2-x) = 1 \)
  Since the left and right limits are equal, the limit exists at \( a = 1 \).

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

• Let \( u = (2x+1)^5 \). Then \( u' = 5(2x+1)^4 \cdot 2 = 10(2x+1)^4 \).
• Let \( v = (x^3-x+1)^4 \). Then \( v' = 4(x^3-x+1)^3 \cdot (3x^2-1) \).

Factor out the common terms: \( 2(2x+1)^4 \) and \( (x^3-x+1)^3 \). $$ \frac{dy}{dx} = 2(2x+1)^4(x^3-x+1)^3 [ 5(x^3-x+1) + (2x+1) \cdot 2(3x^2-1) ] $$ Simplify the expression in the brackets: $$ [ 5x^3 - 5x + 5 + (4x+2)(3x^2-1) ] $$ $$ [ 5x^3 - 5x + 5 + (12x^3 - 4x + 6x^2 - 2) ] $$ $$ [ 17x^3 + 6x^2 - 9x + 3 ] $$

Solution

Part (i): Implicit Differentiation and Tangent Line

1. Find \( \frac{dy}{dx} \): Differentiate \( x^2 + y^2 = 25 \) with respect to \( x \). $$ 2x + 2y \frac{dy}{dx} = 0 $$ $$ 2y \frac{dy}{dx} = -2x $$ $$ \frac{dy}{dx} = \frac{-2x}{2y} = -\frac{x}{y} $$
2. Find Tangent Line at (3, 4):
First, find the slope (m) at (3, 4): $$ m = \frac{dy}{dx} \Big|_{(3,4)} = -\frac{3}{4} $$ Using the point-slope form \( y - y_1 = m(x - x_1) \): $$ y - 4 = -\frac{3}{4}(x - 3) $$ Multiply by 4: $$ 4(y - 4) = -3(x - 3) $$ $$ 4y - 16 = -3x + 9 $$ $$ 3x + 4y = 25 $$

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

• \( f(x) = xe^x \)
• \( f'(x) = (1)e^x + (x)e^x = e^x(1 + x) \)
• \( f''(x) = (e^x)(1+x) + (e^x)(1) = e^x(1 + x + 1) = e^x(2 + x) \)
• \( f'''(x) = (e^x)(2+x) + (e^x)(1) = e^x(2 + x + 1) = e^x(3 + x) \)

The pattern is clear: the n-th derivative adds n to 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 \( x = -1 \): $$ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (ax - b) = a(-1) - b = -a - b $$ $$ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (4x^3 - 12x^2 + 1) = 4(-1)^3 - 12(-1)^2 + 1 = -4 - 12 + 1 = -15 $$ For continuity, we must have: \( -a - b = -15 \) or \( a + b = 15 \).
2. Continuity at \( x = 1 \): $$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (4x^3 - 12x^2 + 1) = 4(1)^3 - 12(1)^2 + 1 = 4 - 12 + 1 = -7 $$ $$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (3) = 3 $$ The left-hand limit (-7) does not equal the right-hand limit (3).

Part (ii): Differentiation
We use the Quotient Rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).

• Let \( u = \sec x \). Then \( u' = \sec x \tan x \).
• Let \( v = 1 + \tan x \). Then \( v' = \sec^2 x \).

$$ \frac{dy}{dx} = \frac{(\sec x \tan x)(1 + \tan x) - (\sec x)(\sec^2 x)}{(1 + \tan x)^2} $$ Expand the numerator: $$ \frac{\sec x \tan x + \sec x \tan^2 x - \sec^3 x}{(1 + \tan x)^2} $$ Factor out \( \sec x \) from the numerator: $$ \frac{\sec x (\tan x + \tan^2 x - \sec^2 x)}{(1 + \tan x)^2} $$ Use the identity \( 1 + \tan^2 x = \sec^2 x \), which means \( \tan^2 x - \sec^2 x = -1 \). $$ \frac{\sec x (\tan x - 1)}{(1 + \tan x)^2} $$

Solution

Part (i): Absolute Extrema

1. Find critical points: \( f'(x) = 12x^3 \).
Set \( f'(x) = 0 \implies 12x^3 = 0 \implies x = 0 \).
2. Test endpoints and critical points: We test the values \( x = -2 \), \( x = 3 \) (endpoints) and \( x = 0 \) (critical point) in the original function.

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

Part (ii-1): Derivative (Product Rule)
\( \frac{d}{dx}(3x^5 \ln x) \)   (Assuming log x is natural log, ln x) $$ = \left[\frac{d}{dx}(3x^5)\right] \cdot (\ln x) + (3x^5) \cdot \left[\frac{d}{dx}(\ln x)\right] $$ $$ = (15x^4)(\ln x) + (3x^5)\left(\frac{1}{x}\right) $$ $$ = 15x^4 \ln x + 3x^4 = 3x^4(5 \ln x + 1) $$

Part (ii-2): Derivative (Quotient Rule)
\( \frac{d}{dx}\left( \frac{x^3}{3x-2} \right) \) $$ = \frac{(3x^2)(3x-2) - (x^3)(3)}{(3x-2)^2} $$ $$ = \frac{9x^3 - 6x^2 - 3x^3}{(3x-2)^2} $$ $$ = \frac{6x^3 - 6x^2}{(3x-2)^2} = \frac{6x^2(x - 1)}{(3x - 2)^2} $$

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

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: $$ f'(x) = \frac{(1)(x^2-x+1) - (x-1)(2x-1)}{(x^2-x+1)^2} $$ Expand the numerator: $$ f'(x) = \frac{(x^2-x+1) - (2x^2 - 3x + 1)}{(x^2-x+1)^2} $$ $$ f'(x) = \frac{x^2 - x + 1 - 2x^2 + 3x - 1}{(x^2-x+1)^2} $$ $$ f'(x) = \frac{-x^2 + 2x}{(x^2-x+1)^2} $$ 2. Find where \( f'(x) = 0 \): $$ -x^2 + 2x = 0 \implies -x(x - 2) = 0 $$ This gives \( x = 0 \) and \( x = 2 \).
3. Find where \( f'(x) \) is undefined:
This occurs if the denominator is zero. \( x^2-x+1 = 0 \).
Using the discriminant (\( b^2-4ac \)): \( (-1)^2 - 4(1)(1) = 1 - 4 = -3 \). Since it's negative, the denominator has no real roots and is never zero.

Part (ii): Differentiation

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

We use the Quotient Rule:

• \( u = \sec x, \quad u' = \sec x \tan x \)
• \( v = 1 + \tan x, \quad v' = \sec^2 x \)

$$ \frac{dy}{dx} = \frac{(\sec x \tan x)(1 + \tan x) - (\sec x)(\sec^2 x)}{(1 + \tan x)^2} $$ Factor out \( \sec x \) in the numerator: $$ \frac{dy}{dx} = \frac{\sec x [\tan x(1 + \tan x) - \sec^2 x]}{(1 + \tan x)^2} $$ $$ \frac{dy}{dx} = \frac{\sec x [\tan x + \tan^2 x - \sec^2 x]}{(1 + \tan x)^2} $$ Using the identity \( \tan^2 x - \sec^2 x = -1 \): $$ \frac{dy}{dx} = \frac{\sec x (\tan x - 1)}{(1 + \tan x)^2} $$

Solution

$$ f(x) = 2x^3 + 3x^2 - 36x $$ $$ f'(x) = 6x^2 + 6x - 36 = 6(x^2 + x - 6) = 6(x+3)(x-2) $$ $$ f''(x) = 12x + 6 = 6(2x+1) $$

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

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

• Interval \( (-\infty, -3) \): Let \( x = -4 \). \( f'(-4) = 6(-1)(-6) = (+) \) (Increasing)
• Interval \( (-3, 2) \): Let \( x = 0 \). \( f'(0) = 6(3)(-2) = (-) \) (Decreasing)
• Interval \( (2, \infty) \): 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.
  \( f(-3) = 2(-27) + 3(9) - 36(-3) = -54 + 27 + 108 = \mathbf{81} \).
• At \( x = 2 \), \( f(x) \) changes from Decreasing to Increasing → Local Minimum.
  \( f(2) = 2(8) + 3(4) - 36(2) = 16 + 12 - 72 = \mathbf{-44} \).

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

• Interval \( (-\infty, -1/2) \): Let \( x = -1 \). \( f''(-1) = 12(-1) + 6 = (-) \) (Concave Down)
• Interval \( (-1/2, \infty) \): Let \( x = 0 \). \( f''(0) = 12(0) + 6 = (+) \) (Concave Up)

The concavity changes at \( x = -1/2 \), so there is an inflection point.
\( f(-1/2) = 2(-1/8) + 3(1/4) - 36(-1/2) = -1/4 + 3/4 + 18 = 1/2 + 18 = 18.5 \)

Solution

$$ f(x) = 2 + 2x^2 - x^4 $$ $$ f'(x) = 4x - 4x^3 = 4x(1 - x^2) = 4x(1-x)(1+x) $$ $$ f''(x) = 4 - 12x^2 = 4(1 - 3x^2) $$

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

(ii) Increasing/Decreasing:

• Interval \( (-\infty, -1) \): 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 \( (1, \infty) \): Let \( x = 2 \). \( f'(2) = 4(2)(1-4) = (-) \) (Decreasing)

(iii) Local Max/Min:

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

(iv) Concavity/Inflection Points:
Set \( f''(x) = 0 \implies 4(1 - 3x^2) = 0 \implies x^2 = 1/3 \implies x = \pm 1/\sqrt{3} \).

• Interval \( (-\infty, -1/\sqrt{3}) \): Let \( x = -1 \). \( f''(-1) = 4 - 12(1) = (-) \) (Concave Down)
• Interval \( (-1/\sqrt{3}, 1/\sqrt{3}) \): Let \( x = 0 \). \( f''(0) = 4 - 0 = (+) \) (Concave Up)
• Interval \( (1/\sqrt{3}, \infty) \): Let \( x = 1 \). \( f''(1) = 4 - 12(1) = (-) \) (Concave Down)

The concavity changes at \( x = \pm 1/\sqrt{3} \).
\( f(\pm 1/\sqrt{3}) = 2 + 2(1/3) - (1/3)^2 = 2 + 2/3 - 1/9 = 18/9 + 6/9 - 1/9 = 23/9 \).

Solution

Part (i): Continuity

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

1. Continuity at \( x = 2 \): $$ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{x^2-4}{x-2} = \lim_{x \to 2^-} (x+2) = 4 $$ $$ \lim_{x \to 2^+} f(x) = f(2) = a(2)^2 - b(2) + 3 = 4a - 2b + 3 $$ Equation 1: \( 4a - 2b + 3 = 4 \)  →  \( 4a - 2b = 1 \)
2. Continuity at \( x = 3 \): $$ \lim_{x \to 3^-} f(x) = f(3) = a(3)^2 - b(3) + 3 = 9a - 3b + 3 $$ $$ \lim_{x \to 3^+} f(x) = 2(3) - a + b = 6 - a + b $$ 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): $$ (10a - 4b) - (8a - 4b) = 3 - 2 $$ $$ 2a = 1 \implies \mathbf{a = 1/2} $$ Substitute \( a = 1/2 \) into (1): \( 4(1/2) - 2b = 1 \implies 2 - 2b = 1 \implies 1 = 2b \implies \mathbf{b = 1/2} \)

Part (ii): Logarithmic Differentiation
Given \( y = x^2 e^{2x} (x^2+1)^4 \)
1. Take the natural log of both sides: $$ \ln y = \ln(x^2) + \ln(e^{2x}) + \ln((x^2+1)^4) $$ $$ \ln y = 2 \ln x + 2x + 4 \ln(x^2+1) $$ 2. Differentiate implicitly w.r.t. \( x \): $$ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x} + 2 + 4 \cdot \frac{1}{x^2+1} \cdot 2x $$ $$ \frac{1}{y} \frac{dy}{dx} = \frac{2}{x} + 2 + \frac{8x}{x^2+1} $$ 3. Solve for \( \frac{dy}{dx} \) by multiplying by \( y \):

Solution

$$ f(x) = x^4 - 2x^3 + 3 $$ $$ f'(x) = 4x^3 - 6x^2 = 2x^2(2x - 3) $$ $$ f''(x) = 12x^2 - 12x = 12x(x - 1) $$

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

(ii) Increasing/Decreasing:

• Interval \( (-\infty, 0) \): Let \( x = -1 \). \( f'(-1) = 2(1)(-5) = (-) \) (Decreasing)
• Interval \( (0, 3/2) \): Let \( x = 1 \). \( f'(1) = 2(1)(-1) = (-) \) (Decreasing)
• Interval \( (3/2, \infty) \): 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.
  \( f(3/2) = (3/2)^4 - 2(3/2)^3 + 3 = 81/16 - 2(27/8) + 3 = 81/16 - 108/16 + 48/16 = \mathbf{21/16} \).

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

• Interval \( (-\infty, 0) \): 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 \( (1, \infty) \): Let \( x = 2 \). \( f''(2) = 12(2)(1) = (+) \) (Concave Up)

Concavity changes at \( x = 0 \) and \( x = 1 \).
\( f(0) = 3 \)
\( f(1) = 1^4 - 2(1)^3 + 3 = 1 - 2 + 3 = 2 \)

Solution

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

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

(ii) Increasing/Decreasing:

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

(iii) Local Max/Min:

• At \( x = \pi/4 \), (Inc → Dec) → Local Maximum.
  \( f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = (1/\sqrt{2}) + (1/\sqrt{2}) = 2/\sqrt{2} = \mathbf{\sqrt{2}} \).
• At \( x = 5\pi/4 \), (Dec → Inc) → Local Minimum.
  \( f(5\pi/4) = \sin(5\pi/4) + \cos(5\pi/4) = (-1/\sqrt{2}) + (-1/\sqrt{2}) = -2/\sqrt{2} = \mathbf{-\sqrt{2}} \).

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

• Interval \( [0, 3\pi/4) \): Let \( x = \pi/2 \). \( f''(\pi/2) = -\sin(\pi/2) - \cos(\pi/2) = -1 = (-) \) (Concave Down)
• Interval \( (3\pi/4, 7\pi/4) \): Let \( x = \pi \). \( f''(\pi) = -\sin(\pi) - \cos(\pi) = 1 = (+) \) (Concave Up)
• Interval \( (7\pi/4, 2\pi] \): Let \( x = 2\pi \). \( f''(2\pi) = -\sin(2\pi) - \cos(2\pi) = -1 = (-) \) (Concave Down)

Inflection points are at \( x = 3\pi/4 \) and \( x = 7\pi/4 \).
\( f(3\pi/4) = \sin(3\pi/4) + \cos(3\pi/4) = (1/\sqrt{2}) + (-1/\sqrt{2}) = 0 \)
\( f(7\pi/4) = \sin(7\pi/4) + \cos(7\pi/4) = (-1/\sqrt{2}) + (1/\sqrt{2}) = 0 \)