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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 2, Problem 2.6.39

Complete the following steps for each function.


c. State the interval(s) of continuity.


f(x)={2x if x<1
x^2+3x if x≥1; a=1

Verified step by step guidance
1
Step 1: Understand the piecewise function. The function f(x) is defined as two separate expressions depending on the value of x. For x < 1, f(x) = 2x. For x \(\geq\) 1, f(x) = x^2 + 3x.
Step 2: Determine the continuity at the point where the function changes, which is at x = 1. For a function to be continuous at a point, the left-hand limit, right-hand limit, and the function value at that point must all be equal.
Step 3: Calculate the left-hand limit as x approaches 1 from the left (x \(\to\) 1^-). Use the expression for x < 1, which is 2x. Evaluate the limit: \(\lim\)_{{x \(\to\) 1^-}} 2x.
Step 4: Calculate the right-hand limit as x approaches 1 from the right (x \(\to\) 1^+). Use the expression for x \(\geq\) 1, which is x^2 + 3x. Evaluate the limit: \(\lim\)_{{x \(\to\) 1^+}} (x^2 + 3x).
Step 5: Check the value of the function at x = 1. Since x = 1 falls into the second piece of the function (x \(\geq\) 1), calculate f(1) using x^2 + 3x. Compare this value with the left-hand and right-hand limits to determine if the function is continuous at x = 1.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Continuity of Functions

A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. For a function to be continuous over an interval, it must be continuous at every point within that interval. Understanding continuity is essential for determining where a function behaves predictably without breaks, jumps, or asymptotes.
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Piecewise Functions

A piecewise function is defined by different expressions based on the input value. In this case, the function f(x) has two distinct expressions: one for x < 1 and another for x ≥ 1. Analyzing piecewise functions requires checking the continuity at the boundaries where the pieces meet, which is crucial for determining the overall continuity of the function.
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Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. To assess continuity at the boundary of a piecewise function, one must evaluate the left-hand limit and the right-hand limit at that point. If both limits exist and are equal to the function's value at that point, the function is continuous there.
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One-Sided Limits
Related Practice
Textbook Question

Determine limxf(x)\(\lim\)_{x\(\rightarrow\]\infty\)}f\(\left\)(x\(\right\)) and limxf(x)\(\lim\)_{x\(\rightarrow\)-\(\infty\)}f\(\left\)(x\(\right\)) for the following functions. Then give the horizontal asymptotes of ff (if any).


f(x)=x6+834x2+3x4+1f\(\left\)(x\(\right\))=\(\frac{\sqrt[3]{x^6+8}\)}{4x^2+\(\sqrt{3x^4+1}\)}

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Textbook Question

Determine the following limits.

lim x→0^− 2 / tan x

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Textbook Question

Find all vertical asymptotes x=ax=a of the following functions. For each value of aa, determine limxa+f(x){\(\displaystyle\]\lim\)_{x\(\to\) a^{+}}}f\(\left\)(x\(\right\)), limxaf(x){\(\displaystyle\]\lim\)_{x\(\to\) a^{-}}}f\(\left\)(x\(\right\)), and limxaf(x){\(\displaystyle\]\lim\)_{x\(\to\) a}}f\(\left\)(x\(\right\)).

f(x)=cos(x)x2+2xf\(\left\)(x\(\right\))=\(\frac{\cos\left(x\right)}{x^2+2x}\)

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Textbook Question

Determine the following limits.

lim x→3 1/ x − 3(1 /√x + 1 − 1/2)

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Textbook Question

Determine the end behavior of the following transcendental functions by analyzing appropriate limits. Then provide a simple sketch of the associated graph, showing asymptotes if they exist. 

f(x)=3exf\(\left\)(x\(\right\))=-3e^{-x}

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Textbook Question

Evaluate limxf(x){\(\displaystyle\[\lim\)_{x\(\to\]\infty\)}{f(x)}} andlimxf(x){\(\displaystyle\]\lim\)_{x\(\to\)-\(\infty\)}{f(x)}}.


f(x)=6ex+203ex+4f\(\left\)(x\(\right\))=\(\frac{6e^{x}\)+20}{3e^{x}+4}

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