Chapter 13: Problem 36
Let
$$h(x)=\left\\{\begin{array}{ll}x & \text { if } x<0 \\ x^{2} & \text { if }
0
Short Answer
Step by step solution
Understanding the Piecewise Function
Evaluating Limit \(\lim_{x \to 0^+} h(x)\)
Evaluating Limit \(\lim_{x \to 0} h(x)\)
Evaluating Limit \(\lim_{x \to 2^-} h(x)\)
Evaluating Limit \(\lim_{x \to 2^+} h(x)\)
Evaluating Limit \(\lim_{x \to 2} h(x)\)
Sketching the Graph of \(h(x)\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits
For \(\lim_{x \to 0^{+}} h(x)\), when \(x\) approaches zero from the positive side, we use \(h(x) = x^2\) since \(0 < x \leq 2\). As \(x\) becomes very small but positive, the function approaches zero. Hence, the limit is 0.
When evaluating \(\lim_{x \to 0} h(x)\), we consider both directions: left and right. From the left \((x \to 0^{-})\), \(h(x)\) equals \(x\), which approaches 0. From the right, already we've noted that \(h(x) = 0\). Since both approaches agree, \(\lim_{x \to 0} h(x) = 0\).
For \(\lim_{x \to 2^{-}} h(x)\), approaching 2 from the left means using \(h(x) = x^2\), and thus the limit dequals 4 since \(2^2 = 4\).
For \(\lim_{x \to 2^{+}} h(x)\), approaching 2 from the right involves \(h(x) = 8 - x\), resulting in the limit of 6, as \(8 - 2 = 6\).
Since \(\lim_{x \to 2^{-}}\) and \(\lim_{x \to 2^{+}}\) don't match, \(\lim_{x \to 2} h(x)\) doesn't exist.
Graphing
- For \(x < 0\), the function is linear: \(y = x\). This graph is a straight line passing through the origin with a 45-degree angle.
- For \(0 < x \leq 2\), the piece \(y = x^2\) is a segment of a parabola. It starts right after \(x = 0\) and ends at \(x = 2\), forming a smooth curve.
- Finally, for \(x > 2\), \(y = 8 - x\) is again linear, but with a negative slope. The line keeps decreasing as \(x\) increases.
Continuity
- At \(x = 0\), the function smoothly transitions from \(y = x\) to \(y = x^2\). The limit from both sides is zero, ensuring continuity.
- At \(x = 2\), there's a different story. Here, the left-hand limit \(\lim_{x \to 2^{-}}\) is 4, while the right-hand limit \(\lim_{x \to 2^{+}}\) is 6. Because these values differ, \(h(x)\) is discontinuous at \(x = 2\).
Parabolic Functions
Parabolas have a characteristic U-shape. For \(h(x)\), this part graphically denotes a steady increase as \(x\) increases from just above 0 to 2. This segment contributes to understanding how \(h(x)\) behaves at different domains.
When graphing a parabola:
- The vertex is a crucial point—here at \((0, 0)\).
- The axis of symmetry for this piece is the y-axis, given \(b = 0\) in \(ax^2 + bx + c\).
- Only part of the parabola is visible (from \(0\) to \(2\)), reflecting its piecewise nature.