Problem 37 Let $$g(x)=\frac{x^{2}+x-6}{|x-2... [FREE SOLUTION]

91影视

Biocalculus Calculus for the Life Sciences

James Stewart

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 2: Problem 37

Let $$g(x)=\frac{x^{2}+x-6}{|x-2|}$$ (a) Find (i) $$\lim _{x \rightarrow 2^{+}} g(x)$$ (ii) $$\lim _{x \rightarrow 2^{-}} g(x)$$ (b) Does $$\lim _{x \rightarrow 2} g(x)$$ exist? (c) Sketch the graph of $$g .$$

Short Answer

Expert verified

(a)(i) 5, (ii) -5; (b) No; (c) Jump discontinuity at x=2.

Step by step solution

Understand the Function

The expression for the function is $ g(x) = \frac{x^2 + x - 6}{|x-2|} $. This means the function is defined as the quotient of a quadratic polynomial and an absolute linear term. Be aware of the behavior of the absolute value, especially around the point where $ x = 2 $.

Evaluate $ \lim_{x \to 2^+} g(x) $

For $ x > 2 $, the expression becomes $ g(x) = \frac{x^2 + x - 6}{x-2} $. Factor the quadratic as $ (x-2)(x+3) $. The factor $ (x-2) $ cancels with the denominator, simplifying to $ x + 3 $. Thus, $ \lim_{x \to 2^+} g(x) = 2 + 3 = 5 $.

Evaluate $ \lim_{x \to 2^-} g(x) $

For $ x < 2 $, the absolute value changes the sign, so $ g(x) = \frac{x^2 + x - 6}{-(x-2)} = -\frac{(x-2)(x+3)}{(x-2)} $. Similarly, their factor cancels, leaving $ -(x + 3) $. Thus, $ \lim_{x \to 2^-} g(x) = -(2 + 3) = -5 $.

Determine the Existence of $ \lim_{x \to 2} g(x) $

Since $ \lim_{x \to 2^+} g(x) = 5 $ and $ \lim_{x \to 2^-} g(x) = -5 $ are not equal, the two-sided limit $ \lim_{x \to 2} g(x) $ does not exist.

Sketch the Graph

To sketch the graph of $ g(x) $, note the behavior around $ x = 2 $. As $ x \to 2^+ $, the graph approaches 5, and as $ x \to 2^- $, it approaches -5, showing a jump discontinuity. For $ x eq 2 $, the graph behaves like $ x + 3 $ for $ x > 2 $ and $ -(x + 3) $ for $ x < 2 $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discontinuity

When we study functions in calculus, we often encounter discontinuities鈥攑oints where a function isn't smooth or doesn't align seamlessly.
At these points, limits play a crucial role in identifying the type of discontinuity. If

The left-hand limit and right-hand limit of a function at a point are not equal
The function takes a sudden jump from one value to another

we have what's known as a jump discontinuity. In our original problem, when considering the limits as

$ x \to 2^+ $, we find that the function approaches 5, while presumably remaining undefined at 2.
$ x \to 2^- $, the function approaches -5.

These unequal limits confirm a jump discontinuity at $ x = 2 $. Thus, the function exhibits a sharp break or jump at this point, underscoring the nonexistence of $ \lim_{x \to 2} g(x) $. This type of discontinuity is distinct as the function itself doesn't merge at $ x = 2 $.

Graph Sketching

Graph sketching is a valuable skill for visualizing functions and understanding their behaviors鈥攎aking abstract math more tangible. To sketch $ g(x) $:

Appreciate the behavior for values around $ x = 2 $.
Note that for $ x > 2 $, the function behaves as $ x+3 $.
For $ x < 2 $, $ g(x) $ mirrors the negative of that expression, as $ -(x+3) $.

Graphical insights offer intuitive understandings鈥攔evealing where $ x = 2 $ produces discord.
This discord is "mapped" as different vertical heights of 5 and -5. This graphical jump reflects the earlier identified discontinuity.
Hence, for students and mathematicians alike, graph sketching serves as a conceptual bridge鈥攂uilding intuition for function limits and behaviors.

Absolute Value Function

The absolute value function, represented by $ |x| $, is a fundamental piece in mathematics denoting the non-negative magnitude of a number. Its properties significantly influence functions involving absolute expressions.

For any real number $ a $, $ |a| $ is $ a $ if $ a \geq 0 $; otherwise, it's $-a$.
It creates two cases in evaluating expressions, adapting based on sign鈥攖his is critical when dealing with expressions like $ |x-2| $.

In the original exercise, the expression $ g(x) = \frac{x^2 + x - 6}{|x-2|} $ results in different expressions for $ x > 2 $ and $ x < 2 $, which are re-organized as:

$ \frac{x^2 + x - 6}{x - 2} $ for $ x > 2 $,
$ \frac{x^2 + x - 6}{-(x - 2)} $ or $ -\frac{x^2 + x - 6}{x - 2} $ for $ x < 2 $.

Such differences are pivotal鈥攊nvoking appropriate expressions for various scenarios.
Grasping this lays foundational blocks for recognizing function transformations and predicting limit behaviors at critical points like discontinuities.

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91影视

Let $$g(x)=\frac{x^{2}+x-6}{|x-2|}$$ (a) Find (i) $$\lim _{x \rightarrow 2^{+}} g(x)$$ (ii) $$\lim _{x \rightarrow 2^{-}} g(x)$$ (b) Does $$\lim _{x \rightarrow 2} g(x)$$ exist? (c) Sketch the graph of $$g .$$

Short Answer

Step by step solution

Understand the Function

Evaluate \( \lim_{x \to 2^+} g(x) \)

Evaluate \( \lim_{x \to 2^-} g(x) \)

Determine the Existence of \( \lim_{x \to 2} g(x) \)

Sketch the Graph

Key Concepts

Discontinuity

Graph Sketching

Absolute Value Function

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