Piecewise Function Evaluation, Graph, and Limit Proof

Given Data

$$f(x) = \begin{cases} 1 + x, & x \leq -1 \ x^2, & x > -1 \end{cases}$$

Evaluate $f(-3)$, $f(-1)$, $f(0)$; sketch graph; prove $\lim_{x\to 0}\frac{|x|}{x}$ does not exist.

Part 1: Evaluations

$f(-3)$: Since $-3 \leq -1$, use $1+x$: $$f(-3) = 1 + (-3) = -2$$

$f(-1)$: Since $-1 \leq -1$, use $1+x$: $$f(-1) = 1 + (-1) = 0$$

$f(0)$: Since $0 > -1$, use $x^2$: $$f(0) = 0^2 = 0$$

Part 2: Graph

Line $y = 1+x$ for $x \leq -1$:

Parabola $y = x^2$ for $x > -1$:

At $x = -1$: left branch gives closed point $(-1, 0)$; right branch approaches open point $(-1, 1)$. The function is discontinuous at $x = -1$ (jump discontinuity of size 1).

f(x)
 4 |                          *
 3 |
 2 |
 1 |              o          *
 0 |            *       *
-1 |          * (line)
-2 |    *   /
   +--+--+--+--+--+--+--+---> x
     -3 -2 -1  0  1  2  3

• Line drawn for $x \le -1$, closed dot at $(-1,0)$
• Parabola drawn for $x > -1$, open dot at $(-1,1)$

Part 3: Prove $\lim_{x\to 0}\frac{|x|}{x}$ does not exist

Recall $|x| = \begin{cases} x, & x > 0 \ -x, & x < 0 \end{cases}$

RHL ($x \to 0^+$): here $|x| = x$: $$\lim_{x\to 0^+}\frac{|x|}{x} = \lim_{x\to 0^+}\frac{x}{x} = 1$$

LHL ($x \to 0^-$): here $|x| = -x$: $$\lim_{x\to 0^-}\frac{|x|}{x} = \lim_{x\to 0^-}\frac{-x}{x} = -1$$

Since $\text{LHL} = -1 \neq 1 = \text{RHL}$, the one-sided limits are unequal.

$$\therefore \lim_{x\to 0}\frac{|x|}{x} \text{ does not exist.} \qquad \blacksquare$$

Verified Solution

Given Data

Part 1: Curve $y = x^2 + 1$

Part 2:

• $z = xy^2 + y^3$
• $x = \sin t$, $y = \cos t$
• Evaluate $\dfrac{dz}{dt}$ at $t = 0$

Part 1: Sketch of $y = x^2 + 1$ (Guidelines)

1. Domain: Polynomial, so domain is $(-\infty, \infty)$.

2. Intercepts:

• $y$-intercept: $x=0 \Rightarrow y = 1$, point $(0,1)$.
• $x$-intercept: $x^2 + 1 = 0 \Rightarrow x^2 = -1$, no real solution. No $x$-intercept.

3. Symmetry: $f(-x) = x^2 + 1 = f(x)$, so symmetric about the $y$-axis (even function).

4. Asymptotes: None (polynomial). As $x \to \pm\infty$, $y \to +\infty$.

5. Increasing/Decreasing: $$\frac{dy}{dx} = 2x$$

• $x < 0$: decreasing
• $x > 0$: increasing
• Critical point at $x = 0$.

6. Local Extremum: $$\frac{d^2y}{dx^2} = 2 > 0$$ So $x = 0$ is a local (and absolute) minimum, $y_{\min} = 1$, point $(0,1)$.

7. Concavity: $y'' = 2 > 0$ everywhere, so concave up everywhere; no inflection points.

8. Table of Values:

9. Sketch: Upward-opening parabola, vertex $(0,1)$, symmetric about $y$-axis.

 y
 5 |   *               *
 4 |
 3 |
 2 |      *         *
 1 |          * (0,1) vertex
   +----+----+----+----+----> x
     -2   -1   0    1    2

Part 2: Find $\dfrac{dz}{dt}$ at $t = 0$

Chain rule: $$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$$

Partial derivatives: $$\frac{\partial z}{\partial x} = y^2, \qquad \frac{\partial z}{\partial y} = 2xy + 3y^2$$

Derivatives w.r.t. $t$: $$\frac{dx}{dt} = \cos t, \qquad \frac{dy}{dt} = -\sin t$$

Combine: $$\frac{dz}{dt} = y^2\cos t + (2xy + 3y^2)(-\sin t)$$

At $t = 0$: $x = \sin 0 = 0$, $y = \cos 0 = 1$, $\cos 0 = 1$, $\sin 0 = 0$.

$$\frac{dz}{dt}\Big|_{t=0} = (1)^2(1) + \big(2(0)(1) + 3(1)^2\big)(-0) = 1 + 0 = 1$$

$$\boxed{\dfrac{dz}{dt}\Big|_{t=0} = 1}$$

Both parts confirmed, with $\frac{dz}{dt}\big|_{t=0} = 1$.

Model Answer

STEP 1 - EXTRACT: Given Data

Part 1 (Area estimation):

• Curve: $y = x^2$
• Bounds: $x = 0$ to $x = 1$
• Method: rectangle method
• Number of sub-intervals: $n = 4$

Part 2 (Particle motion):

• Velocity $v$ at time $t$: the explicit velocity function is MISSING from the question (shown only as "(1)"). The time interval $1 \le t \le 4$ is given.
• Required: (1) displacement, (2) distance travelled over $1 \le t \le 4$.

Missing data note: The velocity expression $v(t)$ was not provided in the readable question. This solution takes $v(t) = \sqrt{t}$, so Part 2 is worked conditionally on that assumption, clearly flagged.

STEP 2 - SOLVE

Part 1: Area under $y = x^2$ on $[0,1]$, $n = 4$

Width of each sub-interval: $$h = \frac{b-a}{n} = \frac{1-0}{4} = 0.25$$

Left-endpoint rule (endpoints $0, 0.25, 0.5, 0.75$):

$$\text{Area} \approx h[f(0)+f(0.25)+f(0.5)+f(0.75)]$$ $$= 0.25(0 + 0.0625 + 0.25 + 0.5625) = 0.25 \times 0.875 = \boxed{0.21875}$$

Right-endpoint rule (endpoints $0.25, 0.5, 0.75, 1.0$): $$= 0.25(0.0625 + 0.25 + 0.5625 + 1.0) = 0.25 \times 1.875 = 0.46875$$

Exact value: $\int_0^1 x^2,dx = \frac{1}{3} \approx 0.3333$.

The estimate depends on which endpoint is chosen; left gives $0.21875$, right gives $0.46875$. Either is a valid "rectangle method" answer.

Part 2 (assuming $v(t) = \sqrt{t}$, unverified)

(1) Displacement: $$s = \int_1^4 \sqrt{t},dt = \frac{2}{3}\left[t^{3/2}\right]_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3} \approx 4.67 \text{ units}$$

(2) Distance: Since $\sqrt{t} \ge 0$ on $[1,4]$, no direction change: $$\text{Distance} = \int_1^4 |\sqrt{t}|,dt = \frac{14}{3} \approx 4.67 \text{ units}$$

Summary

The results are a left-rectangle area of $0.21875$, with displacement and distance both $14/3$. The velocity function for Part 2 is not given in the question, so Part 2 is valid only under the stated assumption.

Initial Value Problem: Definition and Solutions

Given Data

• ODE: $y'' + 4y' - 6y = 0$
• Initial conditions: $y(0) = 1$, $y'(0) = 0$
• Second task: Taylor (Maclaurin) series of $\cos x$ at $x = 0$

Definition of Initial Value Problem

An initial value problem (IVP) is a differential equation together with the values of the unknown function and its derivatives specified at a single point $x_0$ (the initial point). For a second-order ODE:

$$y'' = f(x, y, y'), \quad y(x_0) = y_0, \quad y'(x_0) = y_1$$

The constants $y_0, y_1$ are the initial conditions, and the aim is to find the particular solution satisfying both the ODE and these conditions.

Solving $y'' + 4y' - 6y = 0$

Step 1: Auxiliary Equation

Assume $y = e^{mx}$:

$$m^2 + 4m - 6 = 0$$

Step 2: Roots

$$m = \frac{-4 \pm \sqrt{16 + 24}}{2} = \frac{-4 \pm \sqrt{40}}{2} = -2 \pm \sqrt{10}$$

Real, distinct roots: $m_1 = -2 + \sqrt{10}$, $m_2 = -2 - \sqrt{10}$.

Step 3: General Solution

$$y = C_1 e^{(-2+\sqrt{10})x} + C_2 e^{(-2-\sqrt{10})x}$$

Step 4: Apply Initial Conditions

$y(0) = 1$: $$C_1 + C_2 = 1 \quad (i)$$

Differentiate: $$y' = C_1(-2+\sqrt{10})e^{(-2+\sqrt{10})x} + C_2(-2-\sqrt{10})e^{(-2-\sqrt{10})x}$$

$y'(0) = 0$: $$C_1(-2+\sqrt{10}) + C_2(-2-\sqrt{10}) = 0$$ $$-2(C_1+C_2) + \sqrt{10}(C_1 - C_2) = 0$$

Using $(i)$: $$-2 + \sqrt{10}(C_1 - C_2) = 0 ;\Rightarrow; C_1 - C_2 = \frac{2}{\sqrt{10}} = \frac{\sqrt{10}}{5} \quad (ii)$$

Step 5: Solve

Adding $(i)$ and $(ii)$: $$2C_1 = 1 + \frac{\sqrt{10}}{5} = \frac{5+\sqrt{10}}{5} ;\Rightarrow; C_1 = \frac{5+\sqrt{10}}{10}$$

Subtracting: $$2C_2 = 1 - \frac{\sqrt{10}}{5} ;\Rightarrow; C_2 = \frac{5-\sqrt{10}}{10}$$

Particular Solution

$$\boxed{y = \frac{5+\sqrt{10}}{10}, e^{(-2+\sqrt{10})x} + \frac{5-\sqrt{10}}{10}, e^{(-2-\sqrt{10})x}}$$

Taylor (Maclaurin) Series of $\cos x$ at $x = 0$

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

Substituting:

$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$

$$\boxed{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}, x^{2n}}$$

Verified results: roots $-2 \pm \sqrt{10}$, constants $\frac{5\pm\sqrt{10}}{10}$, and the standard cosine series.

Rolle's Theorem: Statement and Verification

Given Data

• Function: $f(x) = x^2 - 9$
• Interval: $[a, b] = [-3, 3]$

Statement of Rolle's Theorem

If a function $f$ satisfies:

1. $f$ is continuous on the closed interval $[a, b]$,
2. $f$ is differentiable on the open interval $(a, b)$,
3. $f(a) = f(b)$,

then there exists at least one point $c \in (a, b)$ such that

$$f'(c) = 0.$$

Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

$f(x) = x^2 - 9$ is a polynomial. Polynomials are continuous everywhere, hence $f$ is continuous on $[-3, 3]$. ✓

Condition 2: Differentiability on $(-3, 3)$

$$f'(x) = 2x$$

This exists for all real $x$, so $f$ is differentiable on $(-3, 3)$. ✓

Condition 3: Equal endpoint values

$$f(-3) = (-3)^2 - 9 = 9 - 9 = 0$$ $$f(3) = (3)^2 - 9 = 9 - 9 = 0$$ $$\therefore f(-3) = f(3) = 0 \checkmark$$

All three conditions hold.

Finding $c$

Set $f'(c) = 0$:

$$2c = 0 \implies c = 0$$

Check: $c = 0 \in (-3, 3)$. ✓

Conclusion

Since all three conditions of Rolle's theorem are satisfied and there exists $c = 0 \in (-3, 3)$ with $f'(c) = 0$, Rolle's theorem is verified for $f(x) = x^2 - 9$ on $[-3, 3]$.

Newton-Raphson Method: Third Approximation for $x^3 - x - 5 = 0$

STEP 1 - EXTRACT (Given data)

• Equation: $f(x) = x^3 - x - 5 = 0$
• Starting value: $x_1 = 1$
• Required: $x_3$ (third approximation)
• Method implied: Newton-Raphson

STEP 2 - SOLVE

Iteration formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \qquad f'(x) = 3x^2 - 1$$

Iteration 1: Find $x_2$

At $x_1 = 1$: $$f(1) = 1 - 1 - 5 = -5$$ $$f'(1) = 3(1)^2 - 1 = 2$$

$$x_2 = 1 - \frac{-5}{2} = 1 + 2.5 = 3.5$$

Iteration 2: Find $x_3$

At $x_2 = 3.5$: $$f(3.5) = (3.5)^3 - 3.5 - 5 = 42.875 - 8.5 = 34.375$$ $$f'(3.5) = 3(3.5)^2 - 1 = 36.75 - 1 = 35.75$$

$$x_3 = 3.5 - \frac{34.375}{35.75} = 3.5 - 0.96154 = 2.53846$$

Result

$$\boxed{x_3 \approx 2.5385}$$

STEP 1 - EXTRACT

Given data:

• Sequence general term: $a_n = \dfrac{n}{\sqrt{5+n}}$
• Task: define a sequence; determine convergence.

All required data present.

STEP 2 - SOLVE

Part 1: Definition of a Sequence

A sequence is a function whose domain is the set of natural numbers $\mathbb{N}$. It is an ordered list of terms:

$$a_1, a_2, a_3, \ldots, a_n, \ldots$$

denoted ${a_n}$, where $a_n$ is the nth (general) term.

Convergence: A sequence ${a_n}$ converges to a limit $L$ if for every $\varepsilon > 0$ there exists $N$ such that

$$|a_n - L| < \varepsilon \quad \text{for all } n > N.$$

If no finite $L$ exists, the sequence diverges.

Part 2: Convergence of $a_n = \dfrac{n}{\sqrt{5+n}}$

$$\lim_{n\to\infty} \frac{n}{\sqrt{5+n}}$$

Step 1: Rewrite. Write $n = \sqrt{n}\cdot\sqrt{n}$ and factor $n$ from the radical:

$$\sqrt{5+n} = \sqrt{n},\sqrt{\frac{5}{n}+1}$$

So

$$\frac{n}{\sqrt{5+n}} = \frac{\sqrt{n}\cdot\sqrt{n}}{\sqrt{n},\sqrt{\dfrac{5}{n}+1}} = \frac{\sqrt{n}}{\sqrt{\dfrac{5}{n}+1}}$$

Step 2: Take the limit. As $n\to\infty$:

• Numerator: $\sqrt{n}\to\infty$
• Denominator: $\sqrt{\dfrac{5}{n}+1}\to\sqrt{0+1}=1$

$$\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt{\dfrac{5}{n}+1}} = \frac{\infty}{1} = \infty$$

Step 3: Conclusion.

$$\lim_{n\to\infty} a_n = \infty \quad(\text{not finite})$$

Therefore the sequence $a_n = \dfrac{n}{\sqrt{5+n}}$ diverges (it grows without bound and does not converge to any real number).