Problem 39 - AP Calculus with 8 Practice Tests 15th

Question

If \(f(x)=\left\\{\begin{array}{ll}\frac{4 x^{2}-4}{x-1}, & x \neq 1 \\ 4, & x=1\end{array},\right.\) which of the following statements is(are) true? I. \(\lim _{x \rightarrow 1} f(x)\) exists II. \(f\) is continuous at \(x=1\) III. \(f\) is differentiable at \(x=1\) (A) none (B) I only (C) I and II only (D) I, II, and III

Step-by-Step Solution

Answer

I only (Option B).

Step 1: Understand the function definition

The function \( f(x) \) is piecewise-defined. For \( x eq 1 \), it is given by \( f(x) = \frac{4x^2 - 4}{x - 1} \). For \( x = 1 \), \( f(x) = 4 \). We need to determine the behavior near \( x = 1 \) and specifically check limits, continuity, and differentiability.

Step 2: Simplify the expression for x ≠ 1

The numerator \( 4x^2 - 4 \) can be factored as \( 4(x^2 - 1) = 4(x + 1)(x - 1) \). Thus, \( f(x) = \frac{4(x + 1)(x - 1)}{x - 1} \) for \( x eq 1 \). Cancelling \( x - 1 \) from the numerator and the denominator (since \( x eq 1 \)), we get \( f(x) = 4(x + 1) \).

Step 3: Evaluate the limit of f(x) as x approaches 1

To find \( \lim_{x \to 1} f(x) \), use the simplified form \( f(x) = 4(x + 1) \). Therefore, \( \lim_{x \to 1} f(x) = \lim_{x \to 1} 4(x + 1) = 4(1 + 1) = 8 \).

Step 4: Determine continuity at x = 1

A function is continuous at a point if the limit as \( x \) approaches that point equals the function's value at that point. Here, \( \lim_{x \to 1} f(x) = 8 \) and \( f(1) = 4 \). Since \( \lim_{x \to 1} f(x) \) does not equal \( f(1) \), \( f \) is not continuous at \( x = 1 \).

Step 5: Evaluate differentiability at x = 1

A function must be continuous at a point to be differentiable there. Since \( f \) is not continuous at \( x = 1 \), it is also not differentiable at \( x = 1 \).

Step 6: Conclusion on the truth of statements I, II, and III

- Statement I: True, because the limit as \( x \to 1 \) exists and equals 8. - Statement II: False, because \( f \) is not continuous at \( x = 1 \). - Statement III: False, because \( f \) is not differentiable at \( x = 1 \). Thus, the correct option is (B) I only.

Key Concepts

Limits

Limits are fundamental in calculus and help us understand how a function behaves as it approaches a certain point. In this exercise, we needed to find the limit of the function \( f(x) \) as \( x \) approaches 1. For \( x eq 1 \), the function simplifies to \( f(x) = 4(x + 1) \). As we approach \( x = 1 \), this simplifies further to \( 4(1 + 1) = 8 \). This shows that even though the function is undefined at \( x = 1 \) due to division by zero in its original form, we can still determine how the function behaves near that point by simplification. Thus, the limit \( \lim_{x \to 1} f(x) \) exists and equals 8. This process highlights the importance of evaluating limits in determining the behavior of functions near points where they may not be directly evaluated.

Continuity

Continuity at a point implies that the function's limit at that point equals the function's actual value at that point. In simpler terms, you shouldn't have to "lift your pencil" when drawing the graph at that point. For the given function, we determined \( \lim_{x \to 1} f(x) = 8 \) through simplification. However, the function is defined such that \( f(1) = 4 \), which is not equal to the limit. This mismatch indicates a jump or a discontinuity at \( x = 1 \). To check for continuity, comparing the limit and the function value is crucial. Hence, since \( \lim_{x \to 1} f(x) eq f(1) \), the function is not continuous at \( x = 1 \). Understanding continuity involves ensuring no breaks in a function's behavior at a given point.

Differentiability

Differentiability is the ability to take a derivative of a function at a particular point. For a function to be differentiable at a point, it must be continuous there. This means differentiability requires a smooth and continuous curve without sharp turns or breaks. In this exercise, since the function is not continuous at \( x = 1 \), one can directly conclude that it is not differentiable at that point either. Differentiability depends heavily on continuity because if there is a jump or a gap, the slope (or the derivative) can't be consistently defined. Therefore, this exercise illustrates that although a function might not be differentiable at a point, understanding the continuity of the function can provide insight into its overall behavior and help to predict differentiability.