Question

in problems 19 - 32, determine whether the function f is continuous at c. 19. f(x)=x² + 1 at c = - 1 20. f(x)=x³ - 5 21. f(x)=\frac{x}{x² + 4} at c = - 2 22. f(x)=\frac{x}{x - 2}

Explanation:

1. Explanation for problem - type determination:

• We are asked to determine whether a function \(f(x)\) is continuous at a given point \(c\). This involves concepts from calculus such as the definition of continuity (\(\lim_{x

ightarrow c}f(x)=f(c)\)), limits of functions, and evaluating function - values. So, we will use the Step - by - Step Format for a problem in the Calculus sub - field of Mathematics.

1. General steps for checking continuity of a function \(y = f(x)\) at \(x = c\):

• Step 1: Calculate \(f(c)\)
• First, we find the value of the function at \(x = c\) by substituting \(x = c\) into the function \(f(x)\).
• **Step 2: Calculate \(\lim_{x

ightarrow c}f(x)\)**

• We may need to use different limit rules such as direct substitution (if the function is a polynomial or a rational function where the denominator is non - zero at \(x = c\)), factoring, or rationalizing depending on the form of the function.
• **Step 3: Compare \(f(c)\) and \(\lim_{x

ightarrow c}f(x)\)**

• If \(\lim_{x

ightarrow c}f(x)=f(c)\), then the function \(f(x)\) is continuous at \(x = c\). If \(\lim_{x
ightarrow c}f(x)
eq f(c)\) or one of them does not exist, then the function is discontinuous at \(x = c\).

Let's take the first problem (\(f(x)=x^{2}+1\) at \(c=-1\)) as an example:

Step1: Calculate \(f(c)\)

Substitute \(x=-1\) into \(f(x)=x^{2}+1\).
\(f(-1)=(-1)^{2}+1=1 + 1=2\)

Step2: Calculate \(\lim_{x

ightarrow - 1}f(x)\)
Since \(f(x)=x^{2}+1\) is a polynomial function, we can use direct substitution.
\(\lim_{x
ightarrow - 1}(x^{2}+1)=(-1)^{2}+1=2\)

Step3: Compare \(f(c)\) and \(\lim_{x

ightarrow c}f(x)\)
Since \(\lim_{x
ightarrow - 1}f(x)=f(-1) = 2\), the function \(f(x)=x^{2}+1\) is continuous at \(x=-1\).

Step1: Calculate \(f(c)\)

Substitute \(x = - 2\) into \(f(x)=\frac{1}{x^{2}+4}\).
\(f(-2)=\frac{1}{(-2)^{2}+4}=\frac{1}{4 + 4}=\frac{1}{8}\)

Step2: Calculate \(\lim_{x

ightarrow - 2}f(x)\)
Since \(y=x^{2}+4\) is a polynomial and \(x^{2}+4
eq0\) for all real \(x\) (when \(x=-2\), \(x^{2}+4 = 8
eq0\)), for the rational function \(f(x)=\frac{1}{x^{2}+4}\), we can use direct substitution.
\(\lim_{x
ightarrow - 2}\frac{1}{x^{2}+4}=\frac{1}{(-2)^{2}+4}=\frac{1}{8}\)

Step3: Compare \(f(c)\) and \(\lim_{x

ightarrow c}f(x)\)
Since \(\lim_{x
ightarrow - 2}f(x)=f(-2)=\frac{1}{8}\), the function \(f(x)=\frac{1}{x^{2}+4}\) is continuous at \(x=-2\).

Answer:

The function \(f(x)=x^{2}+1\) is continuous at \(x = - 1\).

Let's take the third problem (\(f(x)=\frac{1}{x^{2}+4}\) at \(c=-2\)) as another example: