Problem 32
Question
$$ \left|\frac{4 c}{c+4}\right| ;[0,1] $$
Step-by-Step Solution
Verified Answer
The expression is \( \frac{4c}{c+4} \) on [0,1], with a range from 0 to \(\frac{4}{5}\).
1Step 1: Analyze the Expression Inside the Absolute Value
First, consider the expression \( \frac{4c}{c+4} \). We need to examine how this expression behaves numerically over the interval \([0,1]\).
2Step 2: Determine Expression Sign in \([0,1]\)
Check if the expression \( \frac{4c}{c+4} \) changes sign over the interval \([0,1]\). Since \(4c\) is always non-negative and \(c+4\) is positive due to \(c\) being non-negative, \(\frac{4c}{c+4}\) is always non-negative over the interval.
3Step 3: Simplify Absolute Value Expression
Due to \( \frac{4c}{c+4} \) being non-negative, the absolute value does not change the expression. Therefore, \( \left| \frac{4c}{c+4} \right| = \frac{4c}{c+4} \) for \(c\) in \([0,1]\).
4Step 4: Evaluate Expression at Interval Endpoints
Now, evaluate \( \frac{4c}{c+4} \) at the endpoints of the interval: \(c=0\), \( \frac{4 \times 0}{0+4} = 0 \), and \(c=1\), \( \frac{4 \times 1}{1+4} = \frac{4}{5} \).
5Step 5: Conclusion on Range of the Function
Over the interval \([0,1]\), the expression \( \left| \frac{4c}{c+4} \right| \) takes values ranging from 0 to \(\frac{4}{5}\).
Key Concepts
Absolute ValueExpression EvaluationInterval AnalysisFunction Range
Absolute Value
The absolute value of a number is its distance from zero on the number line, regardless of direction. In mathematical terms, the absolute value of a number \( x \), denoted as \( |x| \), is:
- \( x \), if \( x \geq 0 \)
- -\( x \), if \( x < 0 \)
Expression Evaluation
Expression evaluation involves calculating the value of an algebraic expression. Let's consider the expression \( \frac{4c}{c+4} \). This can be evaluated based on the value given for the variable \( c \). Through substitution and simplification, the numerical value of the expression is determined.
When evaluating expressions, it's crucial to pay attention to:
When evaluating expressions, it's crucial to pay attention to:
- The domain of the expression, meaning the values \( c \) can take without the denominator becoming zero.
- Whether the expression is defined over that domain interval.
Interval Analysis
Interval analysis refers to examining how an expression behaves over a specified range (or interval) of input values. The problem examined the behavior of \( \frac{4c}{c+4} \) over the interval \([0,1]\). Understanding how an expression behaves within an interval is crucial when determining the function's output range or when solving inequality problems.
In this exercise, we needed to verify if the expression remained non-negative and continuous over the chosen interval, which dictates no sudden jumps or undefined points. The key steps involved in interval analysis are:
In this exercise, we needed to verify if the expression remained non-negative and continuous over the chosen interval, which dictates no sudden jumps or undefined points. The key steps involved in interval analysis are:
- Checking sign changes within the interval to identify behavior consistency.
- Evaluating endpoints separately to ensure they are included in the function's range.
Function Range
The range of a function is the set of all possible output values it can produce for given inputs. After determining how the function behaves over a specific interval (from interval analysis), finding its range becomes straightforward.
In the problem, we analyzed \( \left| \frac{4c}{c+4} \right| \) to find it behaved from 0 to \( \frac{4}{5} \) over \([0, 1]\). The endpoints \( c=0 \) and \( c=1 \) were used to establish these boundaries:
In the problem, we analyzed \( \left| \frac{4c}{c+4} \right| \) to find it behaved from 0 to \( \frac{4}{5} \) over \([0, 1]\). The endpoints \( c=0 \) and \( c=1 \) were used to establish these boundaries:
- At \( c = 0: \ rac{4 \times 0}{0+4} = 0 \)
- At \( c = 1: \ rac{4 \times 1}{1+4} = \frac{4}{5} \)
Other exercises in this chapter
Problem 32
Find the radius of convergence of $$ \sum_{n=0}^{\infty} \frac{(p n) !}{(n !)^{p}} x^{n} $$ where \(p\) is a positive integer.
View solution Problem 32
Let \(f(x)=(1+x)^{1 / 2}+(1-x)^{1 / 2}\). Find the Maclaurin series for \(f\) and use it to find \(f^{(4)}(0)\) and \(f^{(51)}(0)\).
View solution Problem 32
For the series given in Problems \(27-32\), determine how large \(n\) must be so that using the nth partial sum to approximate the series gives an error of no m
View solution Problem 32
\(\sum_{n=2}^{\infty}\left(1-\frac{1}{n}\right)^{n}\)
View solution