Model Answer

Part 1: Evaluating the Piecewise Function

Given data: $$f(x) = \begin{cases} 1 + x & \text{if } x \leq -1 \ x^2 & \text{if } x > -1 \end{cases}$$

Points to evaluate: $x = -3,\ -1,\ 0$.

a) $f(-3)$

Since $-3 \leq -1$, use the first piece $f(x) = 1 + x$: $$f(-3) = 1 + (-3) = \boxed{-2}$$

b) $f(-1)$

Since $-1 \leq -1$, use the first piece $f(x) = 1 + x$: $$f(-1) = 1 + (-1) = \boxed{0}$$

c) $f(0)$

Since $0 > -1$, use the second piece $f(x) = x^2$: $$f(0) = (0)^2 = \boxed{0}$$

Sketch of the Graph

• For $x \leq -1$: straight line $y = 1 + x$ (slope $1$), ending at the closed point $(-1, 0)$.
• For $x > -1$: parabola $y = x^2$, approaching the open point $(-1, 1)$ from the right and passing through the origin $(0,0)$.

Note: The left piece gives value $0$ at $x=-1$, but the right-branch limit as $x \to -1^+$ is $(-1)^2 = 1$. Hence the two branches do not meet; there is a jump at $x = -1$ (left value $0$, right limiting value $1$).

      y
      |        .
    1 |o       .  (parabola y = x²)
      | \     .
      |  \   .
    0 |---●--*----------- x
      |  / -1  0
      | /
   -2 ●(-3,-2)   (line y = 1+x)

(● closed point, o open point)

Part 2: Types of Discontinuity

A function $f(x)$ is discontinuous at $x = a$ if $$\lim_{x \to a} f(x) \neq f(a),$$ or the limit does not exist, or $f(a)$ is undefined.

1. Removable Discontinuity: $\lim_{x \to a} f(x)$ exists (finite) but either $f(a)$ is undefined or $\lim_{x\to a} f(x) \neq f(a)$. Appears as a hole; can be removed by redefining $f(a)$.

2. Jump Discontinuity: Both one-sided limits exist but are unequal: $$\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x).$$

3. Infinite Discontinuity: At least one one-sided limit is $\pm\infty$ (vertical asymptote).

Part 3: Analyzing $f(x) = \dfrac{x-2}{x^2 - 7x + 10}$

Step 1: Factorize the denominator

$$x^2 - 7x + 10 = (x-2)(x-5)$$

Step 2: Simplify

$$f(x) = \frac{x-2}{(x-2)(x-5)} = \frac{1}{x-5}, \qquad x \neq 2$$

Step 3: Points of discontinuity

The original function is undefined at $x = 2$ and $x = 5$.

At $x = 2$ (Removable): $$\lim_{x \to 2} f(x) = \frac{1}{2-5} = -\frac{1}{3}$$ Limit exists (finite) but $f(2)$ is undefined → removable discontinuity (hole at $\left(2, -\tfrac{1}{3}\right)$).

At $x = 5$ (Infinite): $$\lim_{x \to 5^-}\frac{1}{x-5} = -\infty, \qquad \lim_{x \to 5^+}\frac{1}{x-5} = +\infty$$ → infinite discontinuity (vertical asymptote at $x = 5$).

Conclusion

• Discontinuous at $x = 2$ (removable) and $x = 5$ (infinite).
• Continuous for all $x \in \mathbb{R} \setminus {2, 5}$.

Note on the graph at $x = -1$: The two pieces do not meet there. The parabola branch approaches $y = 1$ as $x \to -1^+$ while the line gives $0$ at $x = -1$, so the graph has a jump at $x = -1$ rather than joining continuously. The numeric results above are unaffected.

Model Answer: Concavity and Analysis of $f(x) = x^4 - 4x^3 + 10$

Given Data

• Function: $f(x) = x^4 - 4x^3 + 10$

1. Definition of Concavity [2 marks]

A function $f$ is concave up on an interval if its graph lies above all its tangent lines on that interval (opens upward, shape $\cup$). Equivalently, $f''(x) > 0$.

A function $f$ is concave down on an interval if its graph lies below all its tangent lines on that interval (opens downward, shape $\cap$). Equivalently, $f''(x) < 0$.

A point where concavity changes is an inflection point.

2. Intervals of Increase/Decrease [3 marks]

First derivative: $$f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$$

Critical points: $4x^2(x-3) = 0 \Rightarrow x = 0,\ x = 3$

Sign analysis:

Result:

• $f$ is decreasing on $(-\infty, 0)$ and $(0, 3)$, i.e., effectively on $(-\infty, 3)$.
• $f$ is increasing on $(3, \infty)$.

3. Concavity Analysis [3 marks]

Second derivative: $$f''(x) = 12x^2 - 24x = 12x(x-2)$$

Possible inflection points: $12x(x-2) = 0 \Rightarrow x = 0,\ x = 2$

Sign analysis:

Result:

• $f$ is concave up on $(-\infty, 0)$ and $(2, \infty)$.
• $f$ is concave down on $(0, 2)$.
• Inflection points at $x = 0$ and $x = 2$.

4. Local Extrema [2 marks]

Critical points: $x = 0$ and $x = 3$.

First Derivative Test:

• At $x = 0$: $f'$ negative to negative $\Rightarrow$ no local extremum.
• At $x = 3$: $f'$ negative to positive $\Rightarrow$ local minimum.

Local minimum value: $$f(3) = 3^4 - 4(3)^3 + 10 = 81 - 108 + 10 = -17$$

Result:

• Local minimum: $f(3) = -17$ at $x = 3$.
• No local maximum exists.

Model Answer: Domain, Range, and Function Transformations

Given Data

• Function: $f(x) = \sqrt{5x + 10}$
• Base function to transform: $y = x^2$
• Transformations: up 1 unit, down 2 units, left 3 units, right 2 units

(a) Domain and Range of $f(x) = \sqrt{5x + 10}$

Domain

The expression under the square root must be non-negative:

$$5x + 10 \geq 0$$ $$5x \geq -10$$ $$x \geq -2$$

Domain: $[-2, \infty)$

Range

A square root output is always non-negative.

• At $x = -2$: $f(-2) = \sqrt{5(-2)+10} = \sqrt{0} = 0$
• As $x \to \infty$: $f(x) \to \infty$

The function increases continuously from $0$ upward.

Range: $[0, \infty)$

(b) Transformations of $y = x^2$

Applying the shifts sequentially:

1. Up 1 unit: $y = x^2 + 1$
2. Down 2 units: $y = x^2 + 1 - 2 = x^2 - 1$
3. Left 3 units (replace $x$ with $x+3$): $y = (x+3)^2 - 1$
4. Right 2 units (replace $x$ with $x-2$): $y = (x+3-2)^2 - 1 = (x+1)^2 - 1$

Final Function

$$\boxed{y = (x+1)^2 - 1}$$

Vertex: $(-1, -1)$ Axis of symmetry: $x = -1$ Opens: upward

Graph Sketch

        y
        |        y=(x+1)^2 - 1
      3 |  \             /
      2 |   \           /
      1 |    \         /
      0 |-----●-------●----- x
        | -2   \     /  0
     -1 |       \_●_/
        |        (-1,-1)
        +----------------------
       -3   -2   -1   0   1

Key points on the final parabola:

• Vertex: $(-1,-1)$
• $x$-intercepts: $(x+1)^2 - 1 = 0 \Rightarrow x = 0$ and $x = -2$, giving $(0,0)$ and $(-2,0)$
• $y$-intercept: at $x=0$, $y = (1)^2 - 1 = 0 \Rightarrow (0,0)$

The vertex $(-1,-1)$ is the minimum point.

Summary of results

• Domain: $[-2, \infty)$, Range: $[0, \infty)$
• Final transformed function: $y = (x+1)^2 - 1$

Model Answer: Asymptotes of $f(x) = x - \sqrt{x^2 + 16}$

Given Data

• Function: $f(x) = x - \sqrt{x^2 + 16}$
• Domain check: $x^2 + 16 > 0$ for all real $x$, so domain is all of $\mathbb{R}$.

(a) Definitions

Horizontal Asymptote: A line $y = L$ is a horizontal asymptote of $y = f(x)$ if $$\lim_{x \to +\infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L.$$

Vertical Asymptote: A line $x = a$ is a vertical asymptote of $y = f(x)$ if $$\lim_{x \to a^{+}} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^{-}} f(x) = \pm\infty.$$

(An oblique/slant asymptote is a line $y = mx + c$, $m \neq 0$, that the curve approaches at infinity.)

(b) Finding Asymptotes

Vertical Asymptotes

Since $x^2 + 16 \geq 16 > 0$ for all $x$, the function is defined and finite everywhere. There is no finite value of $x$ where $f(x)$ becomes unbounded.

No vertical asymptotes.

Behaviour as $x \to +\infty$

Rationalize: $$f(x) = x - \sqrt{x^2 + 16} = \frac{(x - \sqrt{x^2+16})(x + \sqrt{x^2+16})}{x + \sqrt{x^2+16}} = \frac{x^2 - (x^2 + 16)}{x + \sqrt{x^2 + 16}} = \frac{-16}{x + \sqrt{x^2 + 16}}$$

As $x \to +\infty$, denominator $\to +\infty$, so $$\lim_{x \to +\infty} f(x) = 0.$$

Horizontal asymptote: $y = 0$ (to the right).

Behaviour as $x \to -\infty$

Using the rationalized form, as $x \to -\infty$ we have $\sqrt{x^2+16} \to +\infty$ while $x \to -\infty$. Note $\sqrt{x^2+16} \approx |x| = -x$, so $$x + \sqrt{x^2 + 16} \to 0^{+}, \qquad \frac{-16}{x + \sqrt{x^2+16}} \to -\infty.$$

So $\lim_{x \to -\infty} f(x) = -\infty$; no horizontal asymptote on the left.

Check for oblique asymptote as $x \to -\infty$: $$m = \lim_{x \to -\infty}\frac{f(x)}{x} = \lim_{x \to -\infty}\frac{x - \sqrt{x^2+16}}{x}.$$ With $x < 0$, $\sqrt{x^2+16} = -x\sqrt{1 + 16/x^2}$, so $$f(x) = x - (-x)\sqrt{1 + \tfrac{16}{x^2}} = x + x\sqrt{1 + \tfrac{16}{x^2}}.$$ $$\frac{f(x)}{x} = 1 + \sqrt{1 + \tfrac{16}{x^2}} \to 1 + 1 = 2 = m.$$ $$c = \lim_{x \to -\infty}\big(f(x) - 2x\big) = \lim_{x \to -\infty}\big(-x + x\sqrt{1+\tfrac{16}{x^2}}\big) = \lim_{x \to -\infty} x\Big(\sqrt{1+\tfrac{16}{x^2}} - 1\Big).$$ Expand: $\sqrt{1+\tfrac{16}{x^2}} - 1 \approx \tfrac{8}{x^2}$, so $x \cdot \tfrac{8}{x^2} = \tfrac{8}{x} \to 0$. Thus $c = 0$, giving the oblique asymptote $y = 2x$ as $x \to -\infty$.

Final Answer

• Vertical asymptote: none.
• Horizontal asymptote: $y = 0$ (as $x \to +\infty$).
• Oblique asymptote: $y = 2x$ (as $x \to -\infty$).

Model Answer: First Order Linear Differential Equations

Part 1: Definition [1 mark]

A first order linear differential equation is one that can be written in the standard form:

$$\frac{dy}{dt} + P(t)y = Q(t)$$

where $P(t)$ and $Q(t)$ are functions of the independent variable $t$ alone. It is called first order because it involves only the first derivative $\frac{dy}{dt}$, and linear because $y$ and $\frac{dy}{dt}$ appear only to the first power (no products, powers, or nonlinear functions of $y$).

Part 2: Solve the IVP [4 marks]

Given data:

• ODE: $t\dfrac{dy}{dt} + 2y = t^3$, $t > 0$
• Initial condition: $y(2) = 1$

Step 1: Standard form

Divide through by $t$ (valid since $t>0$):

$$\frac{dy}{dt} + \frac{2}{t}y = t^2$$

So $P(t) = \dfrac{2}{t}$, $Q(t) = t^2$.

Step 2: Integrating factor

$$\mu(t) = e^{\int \frac{2}{t},dt} = e^{2\ln t} = t^2$$

Step 3: Multiply through

$$t^2\frac{dy}{dt} + 2ty = t^4 \quad\Longrightarrow\quad \frac{d}{dt}\big(t^2 y\big) = t^4$$

Step 4: Integrate

$$t^2 y = \frac{t^5}{5} + C$$ $$y = \frac{t^3}{5} + \frac{C}{t^2}$$

Step 5: Apply $y(2) = 1$

$$1 = \frac{2^3}{5} + \frac{C}{2^2} = \frac{8}{5} + \frac{C}{4}$$ $$\frac{C}{4} = 1 - \frac{8}{5} = -\frac{3}{5} \quad\Longrightarrow\quad C = -\frac{12}{5}$$

Step 6: Particular solution

$$\boxed{y = \frac{t^3}{5} - \frac{12}{5t^2} = \frac{t^5 - 12}{5t^2}}$$

Check: At $t=2$: $y = \dfrac{32 - 12}{5\cdot 4} = \dfrac{20}{20} = 1.$ ✓