Chapter 3: Problem 48
Find all points \((x, y)\) on the graph of \(g(x)=\frac{1}{3} x^{3}-\frac{3}{2} x^{2}+1\) with tangent lines parallel to the line \(8 x-2 y=1\)
Short Answer
Expert verified
Points are \((4, \frac{43}{3})\) and \((-1, -\frac{5}{6})\).
Step by step solution
01
Determine the Slope of Parallel Line
First, rewrite the equation of the line \(8x - 2y = 1\) in slope-intercept form, \(y = mx + b\), to find the slope. To do this, solve for \(y\): \(-2y = -8x + 1\), then divide everything by -2 to get \(y = 4x - \frac{1}{2}\). This shows the slope \(m = 4\).
02
Find the Derivative of \(g(x)\)
Calculate the derivative \(g'(x)\) to find the slope of the tangent line at any point \(x\). The function is \(g(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 + 1\). Differentiating, we get \(g'(x) = x^2 - 3x\).
03
Set the Derivative Equal to the Slope
Set \(g'(x) = 4\) because we need the tangent line to have the same slope as \(y = 4x - \frac{1}{2}\). So, solve \(x^2 - 3x = 4\), leading to the equation \(x^2 - 3x - 4 = 0\).
04
Solve the Quadratic Equation
To solve \(x^2 - 3x - 4 = 0\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -3\), and \(c = -4\). This gives \(x = \frac{3 \pm \sqrt{9 + 16}}{2}\), resulting in \(x = 4\) or \(x = -1\).
05
Find Corresponding \(y\)-Coordinates
Substitute \(x = 4\) and \(x = -1\) into \(g(x)\) to find the corresponding \(y\)-coordinates. For \(x = 4\), \(g(4) = \frac{1}{3} \cdot 64 - \frac{3}{2} \cdot 16 + 1 = \frac{64}{3} - 24 + 1 = \frac{43}{3}\). For \(x = -1\), \(g(-1) = \frac{1}{3} \cdot (-1)^3 - \frac{3}{2} \cdot (-1)^2 + 1 = -\frac{1}{3} - \frac{3}{2} + 1 = -\frac{5}{6}\).
06
List the Points
The points on the graph of \(g(x)\) where the tangent lines are parallel to \(8x - 2y = 1\) are \((4, \frac{43}{3})\) and \((-1, -\frac{5}{6})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This type of equation graphs a parabola on a coordinate plane, which can open upwards or downwards depending on the sign of \(a\).
- If \(a\) is positive, the parabola opens upwards.
- If \(a\) is negative, the parabola opens downwards.
- Factoring, when the equation can be broken into simpler expressions that multiply to form the quadratic.
- Completing the square, which involves rearranging the equation to make it easier to solve.
- The quadratic formula, which is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula works for any quadratic equation and provides the solutions for \(x\) in terms of \(a\), \(b\), and \(c\).
Derivative of a Function
The derivative of a function represents the rate at which the function's value changes as its input changes. It's a fundamental concept in calculus, providing the slope of the tangent line to the curve of the function at any given point. For a function \(f(x)\), the derivative \(f'(x)\) can be understood as a way to find the instantaneous rate of change or the slope of the function at a specific point.
- To differentiate a polynomial like \(f(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 + 1\), apply the power rule, which states that \((ax^n)' = anx^{n-1}\).
- Thus, the derivative \(g'(x)\) of our function \(g(x)\) is \(x^2 - 3x\).
Slope-Intercept Form
The slope-intercept form of a linear equation is expressed as \(y = mx + b\), where \(m\) represents the slope of the line, and \(b\) is the y-intercept, which is the point where the line crosses the y-axis. This form is particularly useful because it allows us to quickly identify the slope and y-intercept of a line just by looking at the equation.
- The slope \(m\) indicates how steep the line is and the direction it tilts.
- A positive \(m\) means the line rises as it moves from left to right.
- A negative \(m\) means the line falls as it moves from left to right.
- The y-intercept \(b\) shows where the line meets the y-axis when \(x = 0\).