Problem 7 - Calculus for Biology and Medicine

Question

Let $$ f(x)=\left\\{\begin{array}{cc} \frac{x^{2}-9}{x-3} & \text { if } x \neq 3 \\ a & \text { if } x=3 \end{array}\right. $$ Which value must you assign to \(a\) so that \(f(x)\) is continuous at \(x=3 ?\)

Step-by-Step Solution

Answer

Assign \(a = 6\) to make \(f(x)\) continuous at \(x=3\).

Step 1: Understand Continuity

For a function to be continuous at a point \(x=c\), three conditions must be satisfied: (1) \(f(c)\) is defined, (2) \(\lim\_{x \to c} f(x)\) exists, and (3) \(\lim\_{x \to c} f(x) = f(c)\). In this problem, we need to ensure these conditions hold true at \(x=3\).

Step 2: Simplify the Function for \(x \neq 3\)

Since we are given \(f(x) = \frac{x^2 - 9}{x-3}\), notice that \(x^2 - 9\) can be factored into \((x-3)(x+3)\). Therefore, \(f(x)\) simplifies to \(x+3\) for \(x eq 3\). So, \(f(x) = x + 3\) when \(x eq 3\).

Step 3: Evaluate Limit as \(x\) Approaches 3

To find \(\lim\_{x \to 3} f(x)\), evaluate the limit of the simplified expression \(f(x) = x + 3\) as \(x\) approaches 3. This gives \(\lim\_{x \to 3} (x + 3) = 3 + 3 = 6\).

Step 4: Set \(f(3) = a\) for Continuity

For \(f(x)\) to be continuous at \(x=3\), \(f(3) = a\) must equal \(\lim\_{x \to 3} f(x)\). Thus, set \(a = 6\).

Step 5: Verify Continuity Conditions

Check that all conditions are met: (1) \(f(3) = a = 6\) is defined, (2) \(\lim\_{x \to 3} f(x) = 6\) exists, and (3) \(\lim\_{x \to 3} f(x) = f(3) = 6\). Therefore, \(f(x)\) is continuous at \(x=3\).

Key Concepts

Limits

Limits help us understand the behavior of a function as it gets close to a particular point, even if it doesn't reach it. Imagine moving along a trail towards a peak. The limit tells us what value the function (or the trail height) is heading for as we get infinitely close to a certain point. It's crucial in determining continuity, which requires that a function doesn't break or jump at that point.
In our problem, we aim to make sure that the limit of the function as it approaches 3 is equal to the function's value at 3. This ensures a smooth path or continuous function.
For example, if we simplify the given function to \(f(x) = x + 3\) for \(x eq 3\), when \(x\) approaches 3, the limit becomes \(3 + 3 = 6\). Hence, the limit at \(x = 3\) needs to match \(f(3)\) for continuity.

Piecewise Functions

Piecewise functions are like instructions with specific steps for different situations. They utilize different expressions based on the input value. In this case, the function changes its expression at \(x=3\).
For \(xeq 3\), the function is \(f(x) = \frac{x^2 - 9}{x-3}\). When \(x=3\), the simplified function is undefined unless we account for it using a constant, which we've labeled as \(a\).
Since our goal is to ensure continuity, the value of \(a\) is pivotal. Imagine you are sewing a quilt using different cloth pieces (expressions). The connection between each piece (continuity at the change points) is key to making it seamless. For \(x=3\), \(f(x)\) adopts the value \(a = 6\), thereby ensuring no gaps or discontinuities.

Factoring Polynomials

Factoring polynomials breaks down complicated expressions into simpler, more manageable factors. It’s like taking a complex Lego structure apart into smaller sets so you can easily manipulate or understand them. This technique helps in simplifying expressions and finding limits.
In our case, \(x^2 - 9\) can be expressed as \((x-3)(x+3)\). This factorization transforms the expression \(\frac{x^2 - 9}{x-3}\) into \(x + 3\) for \(x eq 3\). This simplifies the problem and makes it easier to evaluate the limit as \(x\) approaches 3.
By factoring, we remove the indeterminate form and can clearly see the behavior of the function without the division by zero issue. It’s a straightforward yet powerful method to dissect and understand complex polynomial expressions.