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\(15-20\) Explain why the function is discontinuous at the given number a. Sketch the graph of the function. \(f(x)=\left\\{\begin{array}{ll}{e^{x}} & {\text { if } x<0} \\ {x^{2}} & {\text { if } x \geqslant 0}\end{array}\right. \quad a=0\)

Short Answer

Expert verified
The function is discontinuous at \(a=0\) because the left and right limits are unequal.

Step by step solution

01

Understand the Piecewise Function

The given function is a piecewise function with two conditions based on the value of \(x\). For \(x < 0\), the function is \(f(x) = e^x\). For \(x \geq 0\), the function is \(f(x) = x^2\). We need to check the behavior of the function as \(x\) approaches \(a = 0\) from both sides.
02

Calculate the Left-Hand Limit

To find the left-hand limit as \(x\) approaches 0, consider the function part where \(x<0\). Thus, the left-hand limit is \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^x = e^0 = 1\).
03

Calculate the Right-Hand Limit

To find the right-hand limit as \(x\) approaches 0, consider the function part where \(x\geq0\). Thus, the right-hand limit is \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 = 0^2 = 0\).
04

Compare Limits to Check Continuity

For \(f(x)\) to be continuous at \(a=0\), the left-hand limit, right-hand limit, and the function value at \(a=0\) must be equal. We have \(\lim_{x \to 0^-} f(x) = 1\) and \(\lim_{x \to 0^+} f(x) = 0\), and \(f(0) = 0^2 = 0\). Since the left and right limits are not equal, the function is discontinuous at \(a=0\).
05

Sketch the Graph of the Function

1. Plot the graph of \(f(x) = e^x\) for \(x < 0\), which is a decreasing curve approaching 1 as \(x\) approaches 0 from the left.2. Plot the graph of \(f(x) = x^2\) for \(x \geq 0\), which is a parabola starting at 0 at \(x=0\).3. This creates a jump at \(x=0\), demonstrating discontinuity since the value jumps from 1 to 0.

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Key Concepts

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

Piecewise Function
A piecewise function is a type of function defined by different expressions based on intervals of the input value. In other words, the function has a "piece" dedicated to certain sections of the number line. In the given exercise, the function is defined differently for values of \(x\) that are less than 0 and for values that are greater than or equal to 0. For \(x < 0\), the function \(f(x)\) is defined as \(e^x\), an exponential decay function. For \(x \geq 0\), the function is \(f(x) = x^2\), a simple quadratic function that forms a parabola.
These two "pieces" are connected at the point \(a = 0\). Understanding piecewise functions is crucial because each segment can represent very different behaviors and contribute to discontinuities.
Limits
Limits are a fundamental concept in calculus, helping us analyze how functions behave as they approach certain points or infinity. In our exercise, the limit is used to determine the behavior of the piecewise function as \(x\) approaches \(a = 0\).
Calculating a limit involves assessing the value that a function gets closer to as the input approaches a particular value. There are two key limits to consider in this scenario: the left-hand limit (for approaching from the left) and the right-hand limit (for approaching from the right). Getting comfortable with limits allows us to analyze smooth transitions or jumps in a function's graph.
Discontinuity
Discontinuity occurs when there is an abrupt change in the value of a function, creating a "jump" or "break" at a particular point. For the given piecewise function, we check for discontinuity at \(x = 0\) by comparing limits. A function is continuous at a point if the left-hand limit, the right-hand limit, and the function's value at that point are all equal.
In our example, the left-hand limit and right-hand limit at \(x = 0\) are 1 and 0, respectively, leading to a discontinuity. The graph will show a clear jump at this point. Recognizing discontinuities can help in various applications, such as understanding sudden changes in physical systems.
Left-Hand Limit
The left-hand limit of a function as \(x\) approaches a specific point, \(a\), is the value that \(f(x)\) approaches as \(x\) gets close to \(a\) from the left side (values \(x < a\)). For the piecewise function in the exercise, as \(x\) approaches 0 from the left (\(x < 0\)), the function \(f(x) = e^x\) dictates the behavior. The left-hand limit is calculated as \(\lim_{x \to 0^-} e^x = e^0 = 1\).
This limit assists in determining the nature of the function's continuity at \(a = 0\). If the left and right-hand limits are not equal, it indicates a discontinuity.
Right-Hand Limit
The right-hand limit involves approaching a point \(a\) from the right side (values \(x > a\)). It is crucial when evaluating piecewise functions and their potential discontinuities. For our function, as \(x\) approaches 0 from the right, the expression \(f(x) = x^2\) becomes significant. Therefore, the right-hand limit is calculated as \(\lim_{x \to 0^+} x^2 = 0\).
Comparing the right-hand limit with the left-hand limit and the function's value at \(a = 0\) uncovers any discrepancies that denote discontinuity. In this case, the right-hand limit and the function's value at \(0\) are equivalent, but this does not align with the left-hand limit, confirming a jump at \(x = 0\).

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Most popular questions from this chapter

Sketch the graph of the following function and use it to determine the values of a for which \(\lim _{x \rightarrow a}\) \(f(x)\) exists: $$f(x)=\left\\{\begin{array}{ll}{2-x} & {\text { if } x<-1} \\ {x} & {\text { if }-1 \leq x<1} \\ {(x-1)^{2}} & {\text { if } x \geqslant 1}\end{array}\right.$$

The point \(P(1,0)\) lies on the curve \(y=\sin (10 \pi / x)\). $$\begin{array}{l}{\text { (a) If } Q \text { is the point }(x, \sin (10 \pi / x)) \text { , find the slope of the secant }} \\ {\text { line } P Q \text { (correct to four decimal places) for } x=2,1.5,1.4 \text { , }} \\\ {1.3,1.2,1.1,0.5,0.6,0.7,0.8, \text { and } 0.9 . \text { Do the slopes }} \\\ {\text { appear to be approaching a limit? }}\end{array}$$ $$\begin{array}{l}{\text { (b) Use a graph of the curve to explain why the slopes of the }} \\ {\text { secant lines in part (a) are not close to the slope of the tan- }} \\ {\text { gent line at } P .}\end{array}$$ $$\begin{array}{l}{\text { (c) By choosing appropriate secant lines, estimate the slope of }} \\ {\text { the tangent line at } \mathrm{P} .}\end{array}$$

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \(g(\mathrm{x})=\sqrt{1+2 \mathrm{x}}\)

J Use the graph of the function \(f(x)=1 /\left(1+e^{1 / x}\right)\) to state the value of each limit, if it exists. If it does not exist, explain why. $${ (a) }\lim _{x \rightarrow 0^{-}} f(x) \quad \text { (b) } \lim _{x \rightarrow 0^{+}} f(x) \quad \text { (c) } \lim _{x \rightarrow 0} f(x)$$

Determine the infinite limit. $$\lim _{x \rightarrow \pi^{-}} \cot x$$

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