Problem 48 - Calculus Early Transcendentals

Question

Let $$g(x)=\left\\{\begin{array}{ll}{x} & {\text { if } x < 1} \\ {3} & {\text { if } x=1} \\ {2-x^{2}} & {\text { if } 1 < x \leqslant 2} \\ {x-3} & {\text { if } x > 2}\end{array}\right.$$ (a) Evaluate each of the following limits, if it exists. (i) $$\lim _{x \rightarrow 1^{-}} g(x)$$ (ii) $$\lim _{x \rightarrow 1} g(x)$$ (iii) \(g(1)\) (iv) $$\lim _{x \rightarrow 2^{-}} g(x)$$ (v) $$\lim _{x \rightarrow 2^{+}} g(x)$$ (vi) $$\lim _{x \rightarrow 2} g(x)$$ (b) Sketch the graph of \(g .\)

Step-by-Step Solution

Answer

(i) 1, (ii) does not exist, (iii) 3, (iv) -2, (v) -1, (vi) does not exist.

Step 1: Understanding the Piecewise Function

The function \( g(x) \) is defined piecewise with different expressions for different intervals of \( x \):- For \( x < 1 \), \( g(x) = x \).- For \( x = 1 \), \( g(x) = 3 \).- For \( 1 < x \leq 2 \), \( g(x) = 2 - x^2 \).- For \( x > 2 \), \( g(x) = x - 3 \).

Step 2: Evaluate Limit as x Approaches 1 from the Left

To find \( \lim_{x \to 1^-} g(x) \), we consider the expression for \( x < 1 \).The limit becomes \( \lim_{x \to 1^-} x = 1 \).

Step 3: Evaluate Limit as x Approaches 1

Since there is a discontinuity at \( x = 1 \), we check the limits from both sides:- Left-hand limit is \( \lim_{x \to 1^-} g(x) = 1 \).- Right-hand limit for \( x=1 \) isn't needed because the function has a definition at this point.Thus, \( \lim_{x \to 1} g(x) \) does not exist due to different left and right values.

Step 4: Evaluate g(1)

Using the piecewise function definition, for \( x = 1 \), \( g(x) = 3 \). Thus, \( g(1) = 3 \).

Step 5: Evaluate Limit as x Approaches 2 from the Left

For \( \lim_{x \to 2^-} g(x) \), we use the expression for \( 1 < x \leq 2 \):\[ g(x) = 2 - x^2 \]Calculating the limit, \( \lim_{x \to 2^-} (2 - x^2) = 2 - 4 = -2 \).

Step 6: Evaluate Limit as x Approaches 2 from the Right

For \( \lim_{x \to 2^+} g(x) \), we use the expression for \( x > 2 \):\[ g(x) = x - 3 \]Calculating the limit, \( \lim_{x \to 2^+} (x - 3) = 2 - 3 = -1 \).

Step 7: Evaluate Limit as x Approaches 2

Since the left hand limit \( \lim_{x \to 2^-} g(x) = -2 \) and the right hand limit \( \lim_{x \to 2^+} g(x) = -1 \) do not match, \( \lim_{x \to 2} g(x) \) does not exist.

Step 8: Sketch the Graph of g(x)

To sketch the graph:- Start with a diagonal line \( g(x) = x \) for \( x < 1 \).- At \( x = 1 \), there is a point discontinuity jumping to \( g(x) = 3 \).- Between \( 1 < x \leq 2 \), use the curve \( g(x) = 2 - x^2 \), which is a downward opening parabola segment.- For \( x > 2 \), sketch a linear function \( g(x) = x - 3 \) starting just after 2.

Key Concepts

Limits in Calculus

Calculus helps us understand how functions behave as they approach certain points, specifically through the concept of limits. Limits allow us to analyze the behavior of a function as the input value gets close to a specific point. In the case of piecewise functions like the one provided in this exercise, limits help us examine changes in the function's definition at the boundaries of defined intervals.

When evaluating limits, it’s crucial to consider from which direction the input is approaching the point. For instance:

• The left-hand limit, represented by \( \lim_{x \to c^-} f(x) \), checks the behavior as \( x \) approaches \( c \) from the left side.
• The right-hand limit, represented by \( \lim_{x \to c^+} f(x) \), considers the behavior approaching from the right.

A limit exists at a point \( x = c \) if both the left-hand and right-hand limits are equal, regardless of the function's actual value at \( c \). In the exercise, the limit \( \lim_{x \to 1} g(x) \) does not exist due to different limits from the left (1) and the right (undefined due to a jump). Similarly, \( \lim_{x \to 2} g(x) \) does not exist because the left-hand limit (-2) and right-hand limit (-1) do not match.

Discontinuity

Discontinuity in a function occurs when there are certain points where the function is not "continuous." This can happen for several reasons, such as jumps, holes, or vertical asymptotes in the graph. In a piecewise function, discontinuities often appear where different pieces of the function meet.

In the exercise, a key example is at \( x = 1 \):

• Before reaching \( x = 1 \), using \( g(x) = x \), approaching from the left results in a value of 1.
• At \( x = 1 \), the function is explicitly defined as 3, creating a jump discontinuity at this point.

Another discontinuity occurs at \( x = 2 \) because the limits approaching from left (-2) and right (-1) do not agree. Recognizing these discontinuities is essential in graph sketching and finding domain issues within functions.

Graph Sketching

Graph sketching involves plotting the function on a coordinate plane and indicating key features such as intercepts, slopes, and discontinuities. Drawing piecewise functions like \( g(x) \) requires careful consideration of each segment independently and recognizing how they connect at specific boundaries.

Here's how you can sketch \( g(x) \):

• For \( x < 1 \), draw the simple linear function \( g(x) = x \), which is a straight diagonal line.
• At \( x = 1 \), plot a distinct point at \( (1,3) \) since the function value jumps to 3 creating a visible gap from the line \( g(x) = x \).
• Between \( 1 < x \leq 2 \), sketch the curve \( g(x) = 2 - x^2 \). This builds a part of a downward-opening parabola.
• When \( x > 2 \), draw the line \( g(x) = x - 3 \), which extends from the previous curve's ending point without touching it due to the existing discontinuity.

Overall, graph sketching of piecewise functions involves careful attention to transitions at specific values, making sure to illustrate any discontinuities prominently.