Calculus Assignment Guidance: Continuity, Intermediate Value Theorem, and Differentiability

Q1. Sketch the function f and discuss continuity at x = -4 and x = 0

Background

Topic: Continuity of Piecewise Functions

This question tests your understanding of how to analyze the continuity of a function defined by different expressions on different intervals. You are asked to examine both left and right continuity at specific points, and to determine whether the function is continuous or discontinuous at those points.

Key Terms and Formulas:

• Continuity at a point: A function f is continuous at x = a if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.

• Left continuity: $\lim_{x \to a^-} f(x)$

• Right continuity: $\lim_{x \to a^+} f(x)$

• Piecewise function: A function defined by different expressions depending on the value of x.

Step-by-Step Guidance

1. Identify the expressions for f(x) in each interval: $f(x) = |x+4|$ for $x \leq -4$, $f(x) = \sqrt{x+4} - 2$ for $-4 < x < 0$, and $f(x) = x^2 + 2$ for $x \geq 0$.

2. To analyze continuity at $x = -4$, compute the left-hand limit $\lim_{x \to -4^-} f(x)$ and the right-hand limit $\lim_{x \to -4^+} f(x)$ using the appropriate expressions.

3. Check if $f(-4)$ is defined and matches the limits from both sides.

4. Repeat the process for $x = 0$: compute $\lim_{x \to 0^-} f(x)$ and $\lim_{x \to 0^+} f(x)$, and compare with $f(0)$.

5. Discuss whether the function is continuous or discontinuous at each point based on your findings.

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Q2. Use the Intermediate Value Theorem to show existence of solutions

Background

Topic: Intermediate Value Theorem (IVT)

This question tests your ability to use the IVT to prove that certain equations have solutions within a given interval. The IVT states that if a function is continuous on a closed interval [a, b], and takes values f(a) and f(b), then it must take every value between f(a) and f(b) at some point in the interval.

Key Terms and Formulas:

• Intermediate Value Theorem: If f is continuous on [a, b] and $N$ is any number between $f(a)$ and $f(b)$, then there exists $c \in (a, b)$ such that $f(c) = N$.

• Continuous function: A function with no breaks, jumps, or holes in its domain.

Step-by-Step Guidance

1. For part (i), rewrite the equation $\sqrt{4 - x^2} = 2 - \cos(\pi x)$ as $h(x) = \sqrt{4 - x^2} - (2 - \cos(\pi x))$ and look for a sign change in $h(x)$ over an interval.

2. Check the continuity of $h(x)$ on the interval where $\sqrt{4 - x^2}$ is defined (i.e., $-2 \leq x \leq 2$).

3. Evaluate $h(x)$ at two points in the interval to see if $h(x)$ changes sign, which would indicate a solution exists by IVT.

4. For part (ii), define $f(x) = \sin(\pi x) - 4 \log_{10}(x^2 + 1)$ and check for a sign change around $-2$.

5. Verify continuity of $f(x)$ for all real $x$, and evaluate at two points to see if $f(x)$ takes on the value $-2$ somewhere.

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Q3. Use the definition of the derivative to establish existence/non-existence at x = 0 and x = 3

Background

Topic: Definition of the Derivative at a Point

This question tests your ability to use the formal definition of the derivative to determine whether a function is differentiable at specific points, especially for piecewise functions.

Key Terms and Formulas:

• Derivative at a point: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

• Piecewise function: A function defined by different expressions depending on the value of x.

• Differentiability: A function is differentiable at a point if the derivative exists at that point.

Step-by-Step Guidance

1. For $x = 0$, use the definition of the derivative to compute the left and right limits for $g(x)$, considering $g(x) = |4x|$ for $x < 3$.

2. Check if the left and right derivatives at $x = 0$ are equal.

3. For $x = 3$, use the definition of the derivative for both sides: $g(x) = |4x|$ for $x < 3$ and $g(x) = 2x^2/3 + 6$ for $x \geq 3$.

4. Compute the left and right derivatives at $x = 3$ and compare them.

5. Determine if the function is differentiable at $x = 0$ and $x = 3$ based on your calculations.

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Q4. Find values of b and m for differentiability at x = 2

Background

Topic: Differentiability of Piecewise Functions

This question tests your ability to use the definition of the derivative to find parameters that make a piecewise function differentiable at a specific point.

Key Terms and Formulas:

• Derivative at a point: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

• Continuity at a point: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

• Differentiability: Requires both continuity and equal left/right derivatives at the point.

Step-by-Step Guidance

1. Set up the condition for continuity at $x = 2$: $f(2^-) = f(2^+)$, i.e., $2^2 = m \cdot 2 + b$.

2. Set up the condition for differentiability at $x = 2$: $f'(2^-)$ (derivative from the left) must equal $f'(2^+)$ (derivative from the right).

3. Compute $f'(x)$ for $x \leq 2$ and $x > 2$ and set $f'(2^-) = f'(2^+)$.

4. Solve the system of equations for $m$ and $b$ using the continuity and differentiability conditions.

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