Problem 33
Question
Find an equation of each of the lines through the point \((-1,2)\) which is a tangent line to the curve \(y=(x-1) /(x+3)\).
Step-by-Step Solution
Verified Answer
The equation of the tangent line is \(y = \frac{1}{4}x + \frac{9}{4}\) passing through (-1,2).
1Step 1: Find the derivative of the curve
The derivative of a function gives the slope of the tangent line at any point. For the curve given by y = \frac{x-1}{x+3}, use the quotient rule to find the derivative: \(y' = \frac{(x+3) \frac{d}{dx}(x-1) - (x-1) \frac{d}{dx}(x+3)}{(x+3)^2} \). Solving this gives: \( y' = \frac{(x+3)(1) - (x-1)(1)}{(x+3)^2} = \frac{4}{(x+3)^2} \).
2Step 2: Set up the equation of the tangent line
The equation of the tangent line to the curve at any point \((a, f(a))\) is given by: \(y - f(a) = f'(a)(x - a) \). Here, we need to find points \(a\) on the curve that pass through \((-1, 2)\).
3Step 3: Solve for x
Set up the equation with the point given: \(2 = \frac{(x-1)}{(x+3)} \). Cross-multiplying to clear the fraction: \(2(x + 3) = x - 1\).Solving for x, we get: \(2x + 6 = x - 1 \ x = -7\).
4Step 4: Find the slope at x = -7
Use the derivative found in Step 1 to find the slope at \(x = -7\). Plug in \(x = -7\) into \(y' = \frac{4}{(x+3)^2} \): \[ y'(-7) = \frac{4}{(-7+3)^2} = \frac{4}{16} = \frac{1}{4} \].
5Step 5: Write the equation of the tangent line
Using the point-slope form and the point \((-1, 2)\) with slope \(\frac{1}{4}\), the equation of the tangent line is: \(y - 2 = \frac{1}{4}(x + 1) \).
6Step 6: Simplify the equation
Simplify the equation: \(y - 2 = \frac{1}{4}x + \frac{1}{4} \). Adding 2 to both sides: \(y = \frac{1}{4}x + \frac{9}{4} \).
Key Concepts
calculusderivativequotient ruletangent line equation
calculus
Calculus is a branch of mathematics that studies how things change. It's split into two main parts: differential calculus and integral calculus.
These concepts are crucial in solving problems involving motion, growth, and other dynamic systems. In this exercise, we're particularly interested in differential calculus, as we'll be using it to determine the slope of the tangent line to a curve.
- Differential Calculus: Focuses on how functions change. It deals with the concept of a derivative, which represents the rate of change of one quantity with respect to another.
- Integral Calculus: Concerned with accumulation of quantities, such as areas under a curve.
These concepts are crucial in solving problems involving motion, growth, and other dynamic systems. In this exercise, we're particularly interested in differential calculus, as we'll be using it to determine the slope of the tangent line to a curve.
derivative
The derivative of a function provides the slope of the tangent line at any given point on the function's graph. It shows how a function changes as its input changes. Mathematically, the derivative of a function \( f(x) \) at a point \( x \) is defined as:
\[ f'(x) = \frac{dy}{dx} = \text{lim}_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
The process of finding a derivative is called differentiation. In our problem, we need to differentiate the function \(y = \frac{x-1}{x+3} \) to find the slope of the curve at any point. We use the quotient rule to find this derivative, as it involves a fraction with one function in the numerator and another in the denominator.
\[ f'(x) = \frac{dy}{dx} = \text{lim}_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
The process of finding a derivative is called differentiation. In our problem, we need to differentiate the function \(y = \frac{x-1}{x+3} \) to find the slope of the curve at any point. We use the quotient rule to find this derivative, as it involves a fraction with one function in the numerator and another in the denominator.
quotient rule
The quotient rule is a method for finding the derivative of a ratio of two functions. If you have a function \( y = \frac{u(x)}{v(x)} \), then its derivative is given by:
\[ \frac{dy}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]
In other words, you differentiate the numerator and the denominator separately, multiply them appropriately, and then divide by the square of the denominator. In our exercise, we used the quotient rule to derive the function \( y = \frac{(x-1)}{(x+3)} \). Applying this rule, we get:
\[ y' = \frac{(x+3)(1) - (x-1)(1)} {(x+3)^2} = \frac{4}{(x+3)^2} \]
Here, we identified \( u(x) = x-1 \) and \( v(x) = x+3 \), calculated their derivatives, and then applied the quotient rule formula.
\[ \frac{dy}{dx} = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]
In other words, you differentiate the numerator and the denominator separately, multiply them appropriately, and then divide by the square of the denominator. In our exercise, we used the quotient rule to derive the function \( y = \frac{(x-1)}{(x+3)} \). Applying this rule, we get:
\[ y' = \frac{(x+3)(1) - (x-1)(1)} {(x+3)^2} = \frac{4}{(x+3)^2} \]
Here, we identified \( u(x) = x-1 \) and \( v(x) = x+3 \), calculated their derivatives, and then applied the quotient rule formula.
tangent line equation
A tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point. Its slope is equal to the slope of the curve at that point. The tangent line equation at a point \((a, f(a))\) is derived using the point-slope form of the equation of a line:
\[ y - f(a) = f'(a)(x - a) \]
In this exercise, we derived the equation of the tangent line to the curve \( y = \frac{x-1}{x+3} \) at the point where it intersects the given point \((-1,2)\). To do this, we:
Finally, we simplified the equation to get:
\[ y = \frac{1}{4}x + \frac{9}{4} \]
\[ y - f(a) = f'(a)(x - a) \]
In this exercise, we derived the equation of the tangent line to the curve \( y = \frac{x-1}{x+3} \) at the point where it intersects the given point \((-1,2)\). To do this, we:
- Found the derivative (i.e., the slope of the tangent).
- Solved equation for the x-value which passes through the point.
- Used the point-slope form to find the tangent line equation.
Finally, we simplified the equation to get:
\[ y = \frac{1}{4}x + \frac{9}{4} \]
Other exercises in this chapter
Problem 31
Suppose that \(y\) is a function of \(v, v\) is a function of \(u\), and \(u\) is a function of \(x\), and that the derivatives \(D_{v} y, D_{u} v\), and \(D_{x
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Find an equation of the tangent line to the curve \(y=8 /\left(x^{2}+4\right)\) at the point \((2,1)\).
View solution Problem 34
Find an equation of each of the tangent lines to the curve \(3 y=x^{3}-3 x^{2}+6 x+4\), which is parallel to the line \(2 x-y+3=0\)
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If \(f^{\prime}, g^{\prime}, f^{\prime \prime}\), and \(g^{\prime \prime}\) exist and if \(h=f \circ g\), express \(h^{\prime \prime}(x)\) in terms of the deriv
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