Problem 25
Question
In Exercises 25 through 30 , an equation is given. Do the following in each of these problems: (a) Find two functions defined by the equation, and state their domains. (b) Draw a sketch of the graph of each of the functions obtained in part (a). (c) Draw a sketch of the graph of the equation. (d) Find the derivative of each of the functions obtained in part (a) and state the domains of the derivatives. (e) Find \(D_{x} y\) by implicit differentiation from the given equation, and verify that the result so obtained agrees with the results in part (d). (f) Find an equation of each tangent line at the given value of \(x_{1}\). $$ y^{2}=4 x-8 ; x_{1}=3 $$
Step-by-Step Solution
Verified Answer
The two functions are y = √(4x - 8) and y = -√(4x - 8). Their domains are x ≥ 2. Their derivatives are 2/√(4x - 8) and -2/√(4x - 8) respectively. The tangent lines at x=3 are y = x - 1 and y = -x + 1.
1Step 1 - Identify functions from the equation
Given the equation: y^2 = 4x - 8. First, solve for y in terms of x. We get y = ±√(4x - 8). Therefore, the two functions are: 1. y = √(4x - 8) 2. y = -√(4x - 8). Next, state their domains. For the functions to be real, the expression inside the square root must be non-negative. Hence, 4x - 8 ≥ 0 ⇒ x ≥ 2.
2Step 2 - Sketch the graphs of each function
To sketch the graphs of y = √(4x - 8) and y = -√(4x - 8), note that both functions begin at the point (2, 0) and extend rightwards. The first function y = √(4x - 8) is a square root function that lies above the x-axis. The second function y = -√(4x - 8) mirrors the first function and lies below the x-axis.
3Step 3 - Sketch the graph of the equation
The equation y^2 = 4x - 8 represents a sideways parabola opening to the right. The vertex of the parabola is at the point (2, 0). Sketch the parabola accordingly, making sure it reflects both the positive and negative branches of the square root.
4Step 4 - Find derivatives of the functions
For y = √(4x - 8), use the chain rule: y' = 1/2 * (4x - 8)^(-1/2) * 4 = 2 / √(4x - 8). For y = -√(4x - 8), similarly: y' = -1/2 * (4x - 8)^(-1/2) * 4 = -2 / √(4x - 8). The domain of the derivatives is the same as the domain of the original functions, x ≥ 2.
5Step 5 - Verify derivatives using implicit differentiation
Differentiate both sides of the original equation implicitly with respect to x: d/dx(y^2) = d/dx(4x - 8) ⇒ 2y dy/dx = 4 ⇒ dy/dx = 2/y. Then substitute the function values of y: For y = √(4x - 8), dy/dx = 2 / √(4x - 8). For y = -√(4x - 8), dy/dx = -2 / √(4x - 8), which matches the earlier result.
6Step 6 - Finding tangent lines at x = 3
First, find y at x = 3: y = ±√(4*3 - 8) = ±2. For y = 2: m1 = 2 / √(4*3 - 8) = 1. Equation of tangent line is: y - 2 = 1(x - 3) ⇒ y = x - 1. For y = -2: m2 = -2 / √(4*3 - 8) = -1. Equation of tangent line is: y + 2 = -1(x - 3) ⇒ y = -x + 1.
Key Concepts
derivative calculationtangent linefunctions and domainsgraph sketchingchain rule
derivative calculation
When dealing with problems involving derivatives, we start by identifying the functions that are derived from the given equation. For example, in the exercise where the equation is given by \(y^2 = 4x - 8 \), we can first solve for \(y\) in terms of \(x\). This leads to two functions: \(y = \sqrt{4x - 8}\) and \(y = -\sqrt{4x - 8}\).
Next, we use differentiation techniques, such as the chain rule, to find the derivatives of these functions. When using the chain rule, we differentiate the outer function first and then the inner function.
For instance, the derivative of \(y = \sqrt{4x - 8}\) is calculated as follows: \(y' = \frac{1}{2}\cdot (4x - 8)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x - 8}}\). Similarly, for \(y = -\sqrt{4x - 8}\), the derivative is \(y' = \frac{-1}{2}\cdot (4x - 8)^{-1/2} \cdot 4 = \frac{-2}{\sqrt{4x - 8}}\). This process ensures we understand how to handle both positive and negative branches of the function.
Next, we use differentiation techniques, such as the chain rule, to find the derivatives of these functions. When using the chain rule, we differentiate the outer function first and then the inner function.
For instance, the derivative of \(y = \sqrt{4x - 8}\) is calculated as follows: \(y' = \frac{1}{2}\cdot (4x - 8)^{-1/2} \cdot 4 = \frac{2}{\sqrt{4x - 8}}\). Similarly, for \(y = -\sqrt{4x - 8}\), the derivative is \(y' = \frac{-1}{2}\cdot (4x - 8)^{-1/2} \cdot 4 = \frac{-2}{\sqrt{4x - 8}}\). This process ensures we understand how to handle both positive and negative branches of the function.
tangent line
The tangent line to a curve at a given point shows how the function behaves at that exact point. To find the equation of a tangent line, we need two main components: the slope of the tangent line at that point and the coordinates of the point itself.
In the given exercise, the point of interest is when \(x = 3\). Plugging \(x = 3\) into the function \(y = \sqrt{4x - 8}\), we get \(y = 2\). For the function \(y = -\sqrt{4x - 8}\), we get \(y = -2\).
The slopes are obtained from the derivatives found earlier. For \(y = \sqrt{4x - 8}\), the derivative at \(x = 3\) is \(1\). For \(y = -\sqrt{4x - 8}\), the slope is \(-1\).
Using the point-slope form of the equation of a line \(y - y_1 = m(x - x_1)\), we get the tangent lines:
In the given exercise, the point of interest is when \(x = 3\). Plugging \(x = 3\) into the function \(y = \sqrt{4x - 8}\), we get \(y = 2\). For the function \(y = -\sqrt{4x - 8}\), we get \(y = -2\).
The slopes are obtained from the derivatives found earlier. For \(y = \sqrt{4x - 8}\), the derivative at \(x = 3\) is \(1\). For \(y = -\sqrt{4x - 8}\), the slope is \(-1\).
Using the point-slope form of the equation of a line \(y - y_1 = m(x - x_1)\), we get the tangent lines:
- For \(y = \sqrt{4x - 8}\) at \(y = 2\): \(y - 2 = 1(x - 3)\), simplifying to \(y = x - 1\).
- For \(y = -\sqrt{4x - 8}\) at \(y = -2\): \(y + 2 = -1(x - 3)\), simplifying to \(y = -x + 1\).
functions and domains
Understanding functions and their domains is crucial in calculus. The domain is the set of all possible input values (x-values) for which the function is defined.
For the equation \(y^2 = 4x - 8\), solving for \(y\) gives us two functions: \(y = \sqrt{4x - 8}\) and \(y = -\sqrt{4x - 8}\).
To find their domains, we need the expression under the square root to be non-negative. Hence, we solve \(4x - 8 \geq 0\), leading to \(x \geq 2\). This means that the domain for both functions is \([2, \infty)\).
When considering the domains of the derivatives, we must ensure the derivatives are also valid within the same domain. Since both original functions and their derivatives involve the expression \(\sqrt{4x - 8}\), the domain remains the same, \(x \geq 2\).
For the equation \(y^2 = 4x - 8\), solving for \(y\) gives us two functions: \(y = \sqrt{4x - 8}\) and \(y = -\sqrt{4x - 8}\).
To find their domains, we need the expression under the square root to be non-negative. Hence, we solve \(4x - 8 \geq 0\), leading to \(x \geq 2\). This means that the domain for both functions is \([2, \infty)\).
When considering the domains of the derivatives, we must ensure the derivatives are also valid within the same domain. Since both original functions and their derivatives involve the expression \(\sqrt{4x - 8}\), the domain remains the same, \(x \geq 2\).
graph sketching
Sketching graphs is a helpful way to visualize functions and their behaviors.
For the functions \(y = \sqrt{4x - 8}\) and \(y = -\sqrt{4x - 8}\), we start by noting that both functions are derived from the equation \(y^2 = 4x - 8\).
Both functions start at the point \((2, 0)\) and extend rightwards. The graph of \(y = \sqrt{4x - 8}\) lies above the x-axis, resembling half of a parabola opening to the right.
Conversely, \(y = -\sqrt{4x - 8}\) is a mirror image of the first function, lying below the x-axis. When both functions are combined, they form a sideways parabola opening to the right, which is a full visual representation of the equation \(y^2 = 4x - 8\).
Understanding these sketches helps students comprehend how functions behave within their domains.
For the functions \(y = \sqrt{4x - 8}\) and \(y = -\sqrt{4x - 8}\), we start by noting that both functions are derived from the equation \(y^2 = 4x - 8\).
Both functions start at the point \((2, 0)\) and extend rightwards. The graph of \(y = \sqrt{4x - 8}\) lies above the x-axis, resembling half of a parabola opening to the right.
Conversely, \(y = -\sqrt{4x - 8}\) is a mirror image of the first function, lying below the x-axis. When both functions are combined, they form a sideways parabola opening to the right, which is a full visual representation of the equation \(y^2 = 4x - 8\).
Understanding these sketches helps students comprehend how functions behave within their domains.
chain rule
The chain rule is a fundamental technique in differentiation, especially when dealing with composite functions. The chain rule states that to differentiate a composite function, you first differentiate the outer function and then multiply it by the derivative of the inner function.
Consider the function \(y = \sqrt{4x - 8}\). Using the chain rule, the outer function is \(\sqrt{u}\) where \(u = 4x - 8\). The derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\). Then, we multiply by the derivative of the inner function \(4x - 8\), which is 4.
This gives us \(y' = \frac{1}{2}\cdot \frac{1}{\sqrt{4x - 8}} \cdot 4 = \frac{2}{\sqrt{4x - 8}}\).
By applying the chain rule methodically, we simplify the process of finding derivatives of composite functions, leading to accurate and understandable results.
Consider the function \(y = \sqrt{4x - 8}\). Using the chain rule, the outer function is \(\sqrt{u}\) where \(u = 4x - 8\). The derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}}\). Then, we multiply by the derivative of the inner function \(4x - 8\), which is 4.
This gives us \(y' = \frac{1}{2}\cdot \frac{1}{\sqrt{4x - 8}} \cdot 4 = \frac{2}{\sqrt{4x - 8}}\).
By applying the chain rule methodically, we simplify the process of finding derivatives of composite functions, leading to accurate and understandable results.
Other exercises in this chapter
Problem 25
A particle is moving along a straight line according to the given equation of motion, where \(s\) \(\mathrm{ft}\) is the directed distance of the particle from
View solution Problem 25
Given \(f(x)=x^{3}\) and \(g(x)=f\left(x^{2}\right)\). Find: (a) \(f^{\prime}\left(x^{2}\right)\); (b) \(g^{\prime}(x)\).
View solution Problem 26
A particle is moving along a straight line according to the given equation of motion, where \(s\) \(\mathrm{ft}\) is the directed distance of the particle from
View solution Problem 26
Let \(f\) be a function whose domain is the set of all real numbers and \(f(a+b)=f(a) \cdot f(b)\) for all \(a\) and \(b\). Furthermore, suppose that \(f(0)=1\)
View solution