Problem 2 - Calculus Single Variable

Question

Determine at which of the points \(-1,1,\) and 2 the given function \(f\) is continuous. $$ f(x)=\left\\{\begin{array}{cl} 1-x & \text { if } x \leq-1 \\ \sqrt{5-x^{2}} & \text { if }-1

Step-by-Step Solution

Answer

The function is continuous at \(x = -1\) and \(x = 2\).

Step 1: Check Continuity at x = -1

For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point must be equal. At \(x = -1\), we have:\[f(x) = 1-x,\text{ if } x \leq -1\]\(f(-1) = 1 - (-1) = 2\).The left-hand limit as \(x\) approaches \(-1\) is \(\lim_{{x \to -1^-}} f(x) = 2\).The right-hand limit as \(x\) approaches \(-1\) is \(\lim_{{x \to -1^+}} \sqrt{5-x^2} = \sqrt{5-(-1)^2} = \sqrt{4} = 2\).Since the left-hand limit, right-hand limit, and \(f(-1)\) are equal to 2, \(f\) is continuous at \(x = -1\).

Step 2: Check Continuity at x = 1

At \(x = 1\), \[f(x) = \sqrt{5 - x^2},\text{ if } -1 < x < 1\] \[f(x) = 2-\cos(\pi x),\text{ if } x \geq 1\] \(f(1) = 2 - \cos(\pi \cdot 1) = 2 - (-1) = 3\).The left-hand limit as \(x\) approaches \(1\) is \(\lim_{{x \to 1^-}} \sqrt{5-x^2} = \sqrt{5-1^2} = \sqrt{4} = 2\).The right-hand limit as \(x\) approaches \(1\) is \(\lim_{{x \to 1^+}} [2-\cos(\pi x)] = 3\), the same as \(f(1)\).Since the left-hand limit is not equal to the function value at \(x = 1\), \(f\) is not continuous at \(x = 1\).

Step 3: Check Continuity at x = 2

At \(x = 2\), \[f(x) = 2 - \cos(\pi x),\text{ if } x \geq 1\] \(f(2) = 2 - \cos(2\pi) = 2 - 1 = 1\).The left-hand limit as \(x\) approaches \(2\) is \(\lim_{{x \to 2^-}} [2 - \cos(\pi x)] = 1\) since \(2 - \cos(2\pi) = 1\).There is no right-hand limit towards \(2\) from greater values (greater values don't change it), and \(f\) is defined to be \(2 - \cos(\pi x)\) for \(x \geq 1\), so \(f\) is continuous at \(x = 2\).

Key Concepts

Left-Hand Limit

When considering the continuity of a piecewise function at a certain point, the first thing to check is the left-hand limit. The goal is to understand what value the function approaches as it gets closer and closer from the left side of that point. This means we look at values slightly less than the point in question.

• The left-hand limit is expressed as \( \lim_{{x \to c^-}} f(x) \), where \( c \) is the point of interest.
• For instance, at \(x = -1\) in our function, the piece \( f(x) = 1-x \) applies for \( x \leq -1 \). Hence, as \(x\) approaches \(-1\) from the left, \( f(x) \) is approaching 2.

Understanding left-hand limits helps us determine whether the function behaves predictably when approached from the left side. For a function to be continuous, this value should equal both the right-hand limit and the function’s actual value at that point.

Right-Hand Limit

Next up is the right-hand limit. Similar to the left-hand limit, this determines the value that the function approaches as you come closer to the point from the right.

• Mathematically, it's represented as \( \lim_{{x \to c^+}} f(x) \).
• In our function's evaluation at \(x = -1\), notice that to its immediate right we change to the expression \( \sqrt{5 - x^2} \). As \(x\) approaches \(-1\) from the right, this evaluates to 2, the same as the left-hand limit.

Right-hand limits serve to verify the function’s steadiness as we approach from higher values. It’s crucial these limits coincide with the left-hand limit and actual function value for continuity, ensuring there's no sudden skip or jump.

Function Value

Another critical element in determining continuity is the actual value of the function at the point in question. Only when this value aligns with both the left-hand and right-hand limits can we say the function is continuous at that point.

• The function value directly at a point is found by substituting that point into the piece of the function’s expression that applies.
• From our analysis at \(x = 1\), \( f(1) \) evaluates using \( 2 - \cos(\pi x) \) giving us 3. However, this differs from the left-hand limit of 2.

This discrepancy tells us that despite the limits, the actual function shifts its defined behavior at \(x = 1\), breaking the required alignment needed for continuity. When all three values—the left and right-hand limits along with the function value at a point—align, only then is the function steady and seamless, or continuous, at that particular spot.