Chapter 3: Problem 42
Exer. 37-46: (a) Sketch the graph of \(f\). (b) Find the domain \(D\) and range \(R\) of \(f\). (c) Find the intervals on which \(f\) is increasing, is decreasing, or is constant. $$ f(x)=\sqrt{4-x} $$
Short Answer
Expert verified
Domain is \((-\infty, 4]\); Range is \([0, \infty)\); \(f(x)\) decreases over its domain.
Step by step solution
01
Understanding the Function
The function given is \( f(x) = \sqrt{4-x} \). It is a square root function combined with a linear term \( -x \). Square root functions are only defined for non-negative values under the square root.
02
Find the Domain of the Function
To ensure the expression under the square root is non-negative, set the inequality: \[ 4-x \geq 0 \] Solve for \( x \):\[ x \leq 4 \] Thus, the domain \( D \) is \( (-\infty, 4] \).
03
Find the Range of the Function
Since \( f(x) = \sqrt{4-x} \) only takes real non-negative values, we start with the maximum of \( x \), which is 4 (the endpoint), resulting in \( f(4) = 0 \). As \( x \) decreases from 4 to negative infinity, the \( f(x) \) increases without bound. Hence, the range \( R \) of \( f \) is \( [0, \infty) \).
04
Sketch the Graph of the Function
To sketch \( f(x) = \sqrt{4-x} \), note when \( x = 4 \), \( f(x) = 0 \). The graph starts at \((4,0)\) and rises to the left since the square root increases as \( x \) decreases from 4. The graph is a half-parabola opening downwards.
05
Determine the Intervals of Increase, Decrease, or Constancy
Calculate the derivative: \[ f'(x) = \frac{d}{dx}[\sqrt{4-x}] = -\frac{1}{2\sqrt{4-x}} \]Since \( f'(x) < 0 \) for all \( x < 4 \), this means the function \( f(x) \) is always decreasing on its entire domain \((-\infty, 4]\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain and Range
When dealing with square root functions like \( f(x) = \sqrt{4-x} \), it's crucial to understand the concepts of domain and range. The **domain** of a function represents all possible input values (\( x \)-values) that the function can accept without leading to mathematical inconsistencies.
For square root functions, we must ensure that the value under the square root sign is non-negative because the square root of a negative number isn't defined in the set of real numbers.
For square root functions, we must ensure that the value under the square root sign is non-negative because the square root of a negative number isn't defined in the set of real numbers.
- To find the domain of \( f(x) = \sqrt{4-x} \), we solve the inequality \( 4-x \geq 0 \), which simplifies to \( x \leq 4 \).
- This means the domain of \( f \) is \( (-\infty, 4] \).
- The square root function outputs only non-negative numbers, starting at 0 when \( x = 4 \).
- Therefore, the range of \( f \) starts from 0 and extends to positive infinity: \( [0, \infty) \).
Graph Sketching
Sketching the graph of a square root function helps visually understand its behavior. For the function \( f(x) = \sqrt{4-x} \), start by identifying key points and the general shape.
- At \( x = 4 \), the calculation \( f(4) = \sqrt{4-4} = 0 \) gives the point \( (4, 0) \).
- Since the function includes a negative linear term \(-x\), the graph's general motion is upwards as \( x \) decreases from 4.
- The graph never dips below the \( x \)-axis because square roots yield only non-negative outputs.
- As \( x \) approaches negative infinity, \( f(x) \) continues to rise.
Interval Analysis
Interval analysis involves inspecting the behavior of functions over specific sections of the \( x \)-axis.
By differentiating \( f(x) = \sqrt{4-x} \), you can determine whether the function is increasing, decreasing, or constant along its domain.The derivative \( f'(x) = -\frac{1}{2\sqrt{4-x}} \) informs us about the rate of change:
By differentiating \( f(x) = \sqrt{4-x} \), you can determine whether the function is increasing, decreasing, or constant along its domain.The derivative \( f'(x) = -\frac{1}{2\sqrt{4-x}} \) informs us about the rate of change:
- Because \( f'(x) < 0 \) for all \( x < 4 \), it is clear that \( f(x) \) is always decreasing in the domain \((-\infty, 4]\).
- At no point does the derivative equal zero or become positive, which means the function does not have any turning points or increasing intervals.