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Chapter 2: Problem 20

Show that \(f\) is continuous at \(a\). $$ f(x)=\sqrt[3]{x^{2}+2} ; \quad a=-5 $$

Short Answer

Expert verified
The function is continuous at \( a = -5 \) because the limit equals the function value.

Step by step solution

01

Understand the Definition of Continuity

A function \( f \) is continuous at a point \( a \) if \( \lim_{x \to a} f(x) = f(a) \). This means the limit of the function as \( x \) approaches \( a \) must equal the value of the function at \( a \).
02

Calculate \( f(a) \)

To find \( f(a) \), substitute \( a = -5 \) into the function. \[ f(-5) = \sqrt[3]{(-5)^{2} + 2} = \sqrt[3]{25 + 2} = \sqrt[3]{27} = 3 \]
03

Find \( \lim_{x \to a} f(x) \)

We need to find the limit of \( f(x) = \sqrt[3]{x^2 + 2} \) as \( x \to -5 \). Substitute values close to \( x = -5 \) into \( f(x) \) to see if it approaches \( 3 \).
04

Solve the Limit

Calculate the limit directly:\[ \lim_{x \to -5} \sqrt[3]{x^2 + 2} = \sqrt[3]{(-5)^2 + 2} = \sqrt[3]{25 + 2} = \sqrt[3]{27} = 3 \] This shows that the limit is 3.
05

Compare the Limit and Function Value

We calculated \( \lim_{x \to -5} f(x) = 3 \) and \( f(-5) = 3 \). Since these values are equal, \( f(x) \) is continuous at \( x = -5 \).

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Key Concepts

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

Limit of a Function
The limit of a function is a fundamental concept in calculus that describes what a function's output approaches as the input nears a specific value. Consider the function \( f(x) = \sqrt[3]{x^2 + 2} \). To determine if this function is continuous at a specific point, say \( x = -5 \), we first need to compute the limit of this function as \( x \) approaches \(-5\).
  • When we substitute points that are very close to \(-5\) into \( f(x) \), the values of \( f(x) \) should get closer to a certain number, which in this case turns out to be \( 3 \).
  • Thus, calculating \( \lim_{x \to -5} \sqrt[3]{x^2 + 2} = 3 \) confirms what value the function approaches.
This result is essential to proving continuity, as the first step requires confirming that the function aligns at a point from both sides as \( x \) tends to \( a \). Once we determine what this limit is, we proceed to check if it matches the function's actual value at the point, \( f(a) \).
Cube Root Function
The cube root function, symbolized as \( \sqrt[3]{x} \), operates by mapping any real number to its cube root. Cube roots have some unique and useful properties, especially when dealing with functions.
  • The cube root function \( f(x) = \sqrt[3]{x} \) is defined for all real numbers, which means there are no restrictions or places where the function is discontinuous or undefined.
  • This quality is beneficial when evaluating functions like \( f(x) = \sqrt[3]{x^2 + 2} \), as the entire real number line is within its domain, simplifying the analysis of continuity.
  • Calculating a cube root, as in \( \sqrt[3]{27} = 3 \), demonstrates how the function translates the input into an output smoothly and consistently.
By understanding these principles, it becomes clear why cube root functions are able to maintain continuity over their entire domain without interruptions or breaks.
Continuity at a Point
Continuity at a point means that the function behaves predictably near that point, with no jumps, gaps, or oscillations. To show a function \( f \) is continuous at \( a = -5 \), three conditions need to be satisfied:
  • \( f(a) \) is defined, meaning we can compute a specific value when substituting \( a \) into \( f \).
  • The limit \( \lim_{x \to a} f(x) \) exists, calculated by approaching \( a \) from both sides and obtaining a consistent value.
  • The value of the limit equals the value of the function at that point, such as \( \lim_{x \to -5} f(x) = f(-5) = 3 \).
For the function \( f(x) = \sqrt[3]{x^2 + 2} \), we've established all these conditions. By verifying that \( f(-5) \) and the limit as \( x \) approaches \(-5\) are both equal to \( 3 \), we confirm \( f(x) \) is continuous at \( x = -5 \). This process is vital because it ensures the function is graphically seamless and mathematically reliable at that point.

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Most popular questions from this chapter

Verify the intermediate value theorem (2.26) for \(f\) on the stated interval \([a, b]\) by showing that if \(f(a) \leq w \leq f(b),\) then \(f(c)=w\) for some \(c\) in \([a, b].\) $$ f(x)=x^{3}+1: \quad[-1,2] $$

If \(0 \leq f(x) \leq c\) for some real number \(c\), prove that \(\lim _{x \rightarrow 0} x^{2} f(x)=0\).

A telephone company charges 25 cents for the first minute of a long-distance call and 15 cents for each additional minute. (a) Find a piecewise-defined function \(C\) for the total cost of a long- distance call of \(x\) minutes. (b) If \(n\) is an integer greater than 1 , find \(\lim _{x \rightarrow n^{-}} C(x)\) and \(\lim _{x \rightarrow n^{-}} C(x)\)

Verify the intermediate value theorem (2.26) for \(f\) on the stated interval \([a, b]\) by showing that if \(f(a) \leq w \leq f(b),\) then \(f(c)=w\) for some \(c\) in \([a, b].\) $$ f(x)=-x^{3} $$

Use theorems on limits to find the limit, if it exists. $$ \lim _{k \rightarrow 2} \sqrt{3 k^{2}+4} \sqrt[3]{3 k+2} $$
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