a) Examine the continuity or discontinuity of the function f(x) = x^2 16x 4 at x = 4 [3] b) Find dydx if x = 2a(t + 3) and y = a(t^2 + 7) [3]

a) Examine the continuity or discontinuity of the function $f(x)=\frac{x^2-16}{x-4}$ at $x=4$

Step 1: Understanding Continuity

A function $f(x)$ is continuous at a point $x=a$ if the following three conditions are met:

1. $f(a)$ is defined.
2. ${\lim }_{x\to a}f(x)$ exists.
3. ${\lim }_{x\to a}f(x)=f(a)$.

If any of these conditions fail, the function is discontinuous at $x=a$.

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Step 2: Check if $f(4)$ is defined

Given: $$f(x)=\frac{x^2-16}{x-4}$$ At $x=4$: $$f(4)=\frac{4^2-16}{4-4}=\frac{16-16}{0}=\frac{0}{0}$$ This is an indeterminate form (0/0), meaning $f(4)$ is not defined at $x=4$.

Conclusion: The first condition for continuity fails. Thus, $f(x)$ is discontinuous at $x=4$ unless the discontinuity can be removed by simplifying the function.

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Step 3: Simplify the function

Factor the numerator: $$x^2-16=(x-4)(x+4)$$ So, $$f(x)=\frac{(x-4)(x+4)}{x-4}$$ For $x\ne 4$, the $(x-4)$ terms cancel out: $$f(x)=x+4$$ This is the simplified form of $f(x)$.

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Step 4: Check the limit as $x\to 4$

Using the simplified form: $$\underset{x\to 4}{\lim }f(x)=\underset{x\to 4}{\lim }(x+4)=4+4=8$$ The limit exists and equals 8.

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Step 5: Define $f(4)$ to remove the discontinuity

Since the limit exists but $f(4)$ is undefined, we can redefine $f(4)$ to make the function continuous: $$f(4)=8$$ Now, the function becomes: $$f(x)=\{\begin{array}{ll}\frac{x^2-16}{x-4}& \text{if }x\ne 4,\\ 8& \text{if }x=4.\end{array}$$ With this definition, all three conditions for continuity are satisfied:

1. $f(4)=8$ is defined.
2. ${\lim }_{x\to 4}f(x)=8$ exists.
3. ${\lim }_{x\to 4}f(x)=f(4)=8$.

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Step 6: Type of Discontinuity

The original function $f(x)=\frac{x^2-16}{x-4}$ has a removable discontinuity (also called a hole) at $x=4$ because the limit exists but the function is not defined there. After redefinition, the discontinuity is removed.

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b) Find $\frac{dy}{dx}$ if $x=2a(t+3)$ and $y=a(t^2+7)$

Step 1: Understand the Problem

We are given $x$ and $y$ as functions of a parameter $t$: $$x=2a(t+3),y=a(t^2+7)$$ We need to find $\frac{dy}{dx}$, which is the derivative of $y$ with respect to $x$.

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Step 2: Use the Chain Rule for Parametric Differentiation

For parametric equations $x=x(t)$ and $y=y(t)$, the derivative $\frac{dy}{dx}$ is given by: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$

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Step 3: Compute $\frac{dx}{dt}$

Given: $$x=2a(t+3)$$ Differentiate with respect to $t$: $$\frac{dx}{dt}=2a\cdot \frac{d}{dt}(t+3)=2a\cdot 1=2a$$

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Step 4: Compute $\frac{dy}{dt}$

Given: $$y=a(t^2+7)$$ Differentiate with respect to $t$: $$\frac{dy}{dt}=a\cdot \frac{d}{dt}(t^2+7)=a\cdot (2t)=2at$$

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Step 5: Compute $\frac{dy}{dx}$

Using the chain rule: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2at}{2a}=t$$

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Final Answers

a) The function $f(x)=\frac{x^2-16}{x-4}$ is discontinuous at $x=4$ due to an indeterminate form (0/0). However, the discontinuity is removable by redefining $f(4)=8$. After redefinition, the function becomes continuous at $x=4$.

b) $\frac{dy}{dx}=\boxed{t}$