Problem 6
Question
Let $$ f(x)=\sqrt{x}, \quad x \geq 0 . $$ (a) Graph \(y=f(x)\) for \(0 \leq x \leq 6\). (b) For which values of \(x\) is \(y=f(x)\) within \(0.2\) of \(2 ?\) (Hint: Find values of \(x\) such that \(|\sqrt{x}-2|<0.2 .)\) (c) Illustrate your result in (b) on the graph that you obtained in (a).
Step-by-Step Solution
Verified Answer
The values of \( x \) where \( y = \sqrt{x} \) is within 0.2 of 2 are \( 3.24 < x < 4.84 \).
1Step 1: Graphing the function
To graph the function \( y = \sqrt{x} \), we need to calculate the values of \( y \) for \( 0 \leq x \leq 6 \). Calculate a few points: \( (0, 0), (1, 1), (4, 2), (6, \sqrt{6}) \), and then sketch the curve. The curve will start at the origin and increase gradually, making a smooth, gentle upward curve.
2Step 2: Set up inequality for solution
We want to find all \( x \) such that \( |\sqrt{x} - 2| < 0.2 \). This can be rewritten as two separate inequalities: \( \sqrt{x} - 2 < 0.2 \) and \( \sqrt{x} - 2 > -0.2 \).
3Step 3: Solve first inequality
Start with \( \sqrt{x} - 2 < 0.2 \). This implies \( \sqrt{x} < 2.2 \). Squaring both sides, we have \( x < 4.84 \).
4Step 4: Solve second inequality
Now consider \( \sqrt{x} - 2 > -0.2 \). This implies \( \sqrt{x} > 1.8 \). Squaring both sides, we have \( x > 3.24 \).
5Step 5: Combine solutions
The solution to \( |\sqrt{x} - 2| < 0.2 \) is the intersection of \( x < 4.84 \) and \( x > 3.24 \). Thus, \( 3.24 < x < 4.84 \).
6Step 6: Illustrate results on graph
On the graph created in Step 1, highlight the section of the curve where \( 3.24 < x < 4.84 \). This is the part of the graph where the function \( y = \sqrt{x} \) is within \( 0.2 \) of \( y = 2 \), and should be shaded or marked distinctly.
Key Concepts
Graphing functionsInequalitiesSquare root functions
Graphing functions
When we graph functions, we're essentially creating a visual representation of how a function behaves. It lets us see how the values change and relate to each other as we move along the x-axis. For a function like \( y = \sqrt{x} \), the process starts by plotting points. This is because the graph is a collection of all the possible pairs of \( x \) and their corresponding \( y \) values.
- Begin at the point \((0,0)\) because the square root of zero is zero.
- As \( x \) increases, calculate additional points, such as \((1,1)\) since \( \sqrt{1} = 1 \).
- Continue with points like \((4,2)\) and \((6, \sqrt{6})\) to see the trend.
Inequalities
Inequalities are a way to express that one quantity is less than or greater than another. In the context of our problem, we want to determine when the function \( y = \sqrt{x} \) stays within a certain distance from a specified value, like \( y = 2 \). The equation \( |\sqrt{x} - 2| < 0.2 \) means we're interested in all \( x \) values where the output of the function is between 1.8 and 2.2. To do this, we analyze two separate inequalities:
- \( \sqrt{x} - 2 < 0.2 \), which means \( \sqrt{x} < 2.2 \).
- \( \sqrt{x} - 2 > -0.2 \), which results in \( \sqrt{x} > 1.8 \).
Square root functions
Square root functions, such as \( f(x) = \sqrt{x} \), have a unique shape. These functions only take non-negative values for \( x \) because square roots of negative numbers aren't real numbers. The function starts at the origin and grows at a decreasing rate, meaning it rises quickly at first before gradually leveling out as \( x \) increases.
Some properties of square root functions include:
Some properties of square root functions include:
- They only operate in the first quadrant of a graph, where both \( x \) and \( y \) values are non-negative.
- They are continuous and smooth, without any breaks or sharp corners.
- Asymptotically, they approach infinity, meaning they keep increasing but at a slower pace.
Other exercises in this chapter
Problem 5
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow \pi} 3 \cos \frac{x}{4} $$
View solution Problem 6
Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{\sin (\pi x / 2)}{2 x} $$
View solution Problem 6
Use the Intermediate Value Theorem to show that \(\cos x=x\) has a solution in \((0,1)\).
View solution Problem 6
Show that $$ f(x)=\left\\{\begin{array}{cc} \frac{2 x^{2}+x-6}{x+2} & \text { if } x \neq-2 \\ -7 & \text { if } x=-2 \end{array}\right. $$ is continuous at \(x
View solution