91Ó°ÊÓ

For each of the following prompts, give an example of a function that satisfies the stated criteria. A formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why. a. A function \(f\) that is continuous at \(a=2\) but not differentiable at \(a=2\). b. A function \(g\) that is differentiable at \(a=3\) but does not have a limit at \(a=3\). c. A function \(h\) that has a limit at \(a=-2,\) is defined at \(a=-2,\) but is not continuous at \(a=-2\) d. A function \(p\) that satisfies all of the following: \- \(p(-1)=3\) and \(\lim _{x \rightarrow-1} p(x)=2\) $$\text { - } p(0)=1 \text { and } p^{\prime}(0)=0$$ \- \(\lim _{x \rightarrow 1} p(x)=p(1)\) and \(p^{\prime}(1)\) does not exist

Step by step solution

01

Part (a): A function that is continuous at \(a=2\) but not differentiable at \(a=2\)

The absolute value function \( f(x) = |x - 2| \) is a perfect example. This function is continuous at \(x = 2\), but it is not differentiable at \(x = 2\) because the left-hand and right-hand derivatives are not equal.

02

Part (b): A function that is differentiable at \(a=3\) but does not have a limit at \(a=3\)

Such a function does not exist. If a function is differentiable at a point \(x\), it must have a limit at that point. Differentiability implies continuity, and continuity implies the existence of a limit.

03

Part (c): A function that has a limit at \(a=-2\), is defined at \(a=-2\), but is not continuous at \(a=-2\)

Consider the function \( h(x) = \begin{cases} x^2 & \text{if } x eq -2 \ 3 & \text{if } x = -2 \end{cases} \). The limit as \(x\) approaches \(-2\) is \(4\), the function is defined at \(x = -2\) with \( h(-2) = 3 \), but since \( h(-2) eq \lim_{x \to -2} h(x) \), the function is not continuous at \(-2\).

04

Part (d): A function that satisfies multiple criteria

Construct the function piecewise as follows:

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

continuous functions

Continuous functions are those which have no interruptions, jumps, or breaks in their graphs. This means that for a function to be continuous at a point \(a\), the following conditions must be met:

• The function must be defined at \(a\).
• The limit of the function as \(x\) approaches \(a\) must exist.
• The value of the function at \(a\) must equal the limit of the function as \(x\) approaches \(a\).

For example, the absolute value function \(f(x) = |x - 2|\) is continuous at \(x = 2\) because it meets all the above conditions. However, it is not differentiable at \(x = 2\). This makes it a good example of a function that is continuous but not differentiable.

differentiable functions

Differentiable functions not only need to be continuous but also must have a defined derivative at every point in their domain. Differentiability ensures the existence of a unique tangent at every point of the function. Important points to note:

• If a function is differentiable at a point \(a\), it means the function is also continuous at \(a\).
• Differentiability implies a unique slope or rate of change at that point.
• If the function has sharp corners or cusps at any point, it is not differentiable there.

For instance, the function \(g(x)\) being differentiable at \(a=3\) but not having a limit at \(a=3\) is impossible. Differentiability inherently requires the function to have a limit at that point because differentiability implies continuity.

limits and continuity

Understanding limits and the role they play in continuity is essential. A limit helps us investigate the behavior of a function as it gets arbitrarily close to a point, but not necessarily reaching it. For continuity at a point \(a\), the following must be true:

• \(\text{lim}_{x \to a} f(x) = L\): The function approaches a specific value \(L\) as \(x\) gets close to \(a\).
• \(f(a)\) is defined.
• \(\text{lim}_{x \to a} f(x) = f(a)\).

A classic example of a function that has a limit but is not continuous at a point is the piecewise function \(h(x)\) given by:
\[ h(x) = \begin{cases} x^2 & \text{if } x e -2 \, 3 & \text{if } x = -2 \end{cases} \]. This function is defined at \(x = -2\) and the limit as \(x\) approaches \(-2\) exists and equals \(4\), but \(h(-2) = 3\). Since \(h(-2)\) is not equal to the limit, the function is not continuous at \(-2\).

piecewise functions

Piecewise functions are defined by different expressions depending on the input value. They allow us to create functions with specific behaviors at certain points. A piecewise function can meet various criteria by combining multiple distinct functions. For example, the function \(p(x)\) satisfying:

• \(p(-1)=3\) and \(\text{lim}_{x \rightarrow -1} p(x)=2\)
• \(p(0)=1\) and \(p^{\text{prime}}(0)=0\)
• \(\text{lim}_{x \rightarrow 1} p(x)=p(1)\) and \(p^{\text{prime}}(1)\) does not exist.

To achieve the specified properties, we can define each segment matching the criteria. Piecewise functions are excellent for modelling real-world scenarios where different rules apply in different contexts.