Math Problem Statement
xLet g(x) =
x if x < 1 2 if x = 1 2 − x2 if 1 < x ≤ 2 x − 3 if x > 2 . (a) Evaluate each of the following, if it exists. (If an answer does not exist, enter DNE.) (i) lim x→1− g(x) (ii) lim x→1 g(x) (iii) g(1) (iv) lim x→2− g(x) (v) lim x→2+ g(x) (vi) lim x→2 g(x) (b) Sketch the graph of g.
The x y coordinate plane is given. A function comprised of two lines, one curve, and one point is on the graph. The point occurs at (1, 2). The first line enters the window in the third quadrant, goes up and right, passes through the origin, and ends at the open point (1, 1). The curve begins at the open point (1, 1), goes down and right becoming more steep, and ends at the closed point (2, −2). The second line begins at the open point (2, 1), goes down and right, crosses the x-axis at x = 3, and exits the window in the fourth quadrant.
The x y coordinate plane is given. A function comprised of two lines and one point is on the graph. The point occurs at (1, 2). The first line enters the window in the third quadrant, goes up and right, passes through the origin, and ends at the open point (1, 1). The second line begins at the open point (1, −2), goes up and right, crosses the x-axis at x = 3, and exits the window in the first quadrant.
The x y coordinate plane is given. A function comprised of two lines and one curve is on the graph. The first line enters the window in the second quadrant, goes horizontally right, crosses the y-axis at y = 1, and ends at the open point (1, 1). The curve begins at the open point (1, 1), goes down and right becoming more steep, and ends at the closed point (2, −2). The second line begins at the open point (2, −1), goes up and right, crosses the x-axis at x = 3, and exits the window in the first quadrant.
The x y coordinate plane is given. A function comprised of two lines, one curve, and one point is on the graph. The point occurs at (1, 2). The first line enters the window in the third quadrant, goes up and right, passes through the origin, and ends at the open point (1, 1). The curve begins at the open point (1, 1), goes down and right becoming more steep, and ends at the closed point (2, −2). The second line begins at the open point (2, −1), goes up and right, crosses the x-axis at x = 3, and exits the window in the first quadrant.
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Solution
Let's go step by step to solve each part of the problem. The function is defined as:
\begin{cases} x & \text{if } x < 1 \\ 2 & \text{if } x = 1 \\ 2 - x^2 & \text{if } 1 < x \leq 2 \\ x - 3 & \text{if } x > 2 \end{cases}$$ ### (a) Evaluate each of the following: #### (i) $$\lim_{x \to 1^-} g(x)$$ This is the limit as $$x$$ approaches 1 from the left (values less than 1). For $$x < 1$$, the function is $$g(x) = x$$. So, $$\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} x = 1$$ #### (ii) $$\lim_{x \to 1} g(x)$$ To evaluate $$\lim_{x \to 1} g(x)$$, we need to check both the left-hand limit and the right-hand limit at $$x = 1$$. - The left-hand limit $$\lim_{x \to 1^-} g(x) = 1$$, as calculated above. - For the right-hand limit, $$g(x) = 2 - x^2$$ for $$1 < x \leq 2$$, so: $$\lim_{x \to 1^+} g(x) = 2 - (1)^2 = 1$$ Since both the left-hand and right-hand limits are equal, the two-sided limit exists, and: $$\lim_{x \to 1} g(x) = 1$$ #### (iii) $$g(1)$$ From the definition of the function, we know that $$g(1) = 2$$. #### (iv) $$\lim_{x \to 2^-} g(x)$$ As $$x$$ approaches 2 from the left, $$g(x) = 2 - x^2$$ for $$1 < x \leq 2$$, so: $$\lim_{x \to 2^-} g(x) = 2 - (2)^2 = 2 - 4 = -2$$ #### (v) $$\lim_{x \to 2^+} g(x)$$ As $$x$$ approaches 2 from the right, $$g(x) = x - 3$$ for $$x > 2$$, so: $$\lim_{x \to 2^+} g(x) = 2 - 3 = -1$$ #### (vi) $$\lim_{x \to 2} g(x)$$ To evaluate $$\lim_{x \to 2} g(x)$$, we need to compare the left-hand limit and the right-hand limit. - Left-hand limit: $$\lim_{x \to 2^-} g(x) = -2$$ - Right-hand limit: $$\lim_{x \to 2^+} g(x) = -1$$ Since the left-hand and right-hand limits are not equal, the limit does not exist (DNE). ### Summary of Results: (i) $$\lim_{x \to 1^-} g(x) = 1$$ (ii) $$\lim_{x \to 1} g(x) = 1$$ (iii) $$g(1) = 2$$ (iv) $$\lim_{x \to 2^-} g(x) = -2$$ (v) $$\lim_{x \to 2^+} g(x) = -1$$ (vi) $$\lim_{x \to 2} g(x) = \text{DNE}$$ --- ### (b) Sketch the graph of $$g(x)$$ The function consists of four parts: 1. For $$x < 1$$, the function is linear: $$g(x) = x$$. - This is a line passing through the origin and has an open point at $$(1, 1)$$. 2. At $$x = 1$$, the value is $$g(1) = 2$$, which is a single point. 3. For $$1 < x \leq 2$$, the function is $$g(x) = 2 - x^2$$, a downward-opening parabola. - The parabola starts at $$(1, 1)$$ (open point) and ends at $$(2, -2)$$ (closed point). 4. For $$x > 2$$, the function is linear: $$g(x) = x - 3$$. - This is a line starting at $$(2, -1)$$ (open point) and has a slope of 1. Would you like me to generate a graph of this function or elaborate on any part of the solution? --- ### Five Related Questions: 1. What happens if the definition of $$g(x)$$ at $$x = 1$$ is changed to $$g(1) = 1$$? 2. How do we compute limits for piecewise functions with different behaviors on either side of the point? 3. What is the significance of continuity at a point? Is $$g(x)$$ continuous at $$x = 1$$? 4. What kind of discontinuities does $$g(x)$$ have, if any? 5. How would the graph of $$g(x)$$ change if the quadratic term in the interval $$(1, 2]$$ was changed? ### Tip: When evaluating limits for piecewise functions, always check the left-hand and right-hand limits separately to determine if the overall limit exists.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Continuity
Formulas
lim x→c⁻ f(x) = L (left-hand limit)
lim x→c⁺ f(x) = L (right-hand limit)
Piecewise function definition
Theorems
Limit Laws
Continuity Definition
Conditions for a Limit to Exist
Suitable Grade Level
Undergraduate Calculus or AP Calculus