Problem 62
Question
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 \(8 x-2 y=1\).
Step-by-Step Solution
Verified Answer
The points are \((4, \frac{5}{3})\) and \((-1, -\frac{5}{6})\).
1Step 1: Find the slope of the given line
The equation of the line is given by \( 8x - 2y = 1 \). To find the slope, first rewrite it in slope-intercept form \( y = mx + b \). Start by isolating \( y \): \( 2y = 8x - 1 \), then \( y = 4x - \frac{1}{2} \). The slope \( m \) of the line is \( 4 \).
2Step 2: Differentiate the given function
The function is \( g(x) = \frac{1}{3} x^3 - \frac{3}{2} x^2 + 1 \). Calculate the derivative \( g'(x) \) to find the slope of the tangent line at any point. The derivative is \( g'(x) = x^2 - 3x \).
3Step 3: Set the derivative equal to the given slope
We want the tangent line to be parallel to the given line, meaning the slopes should be equal. Set the derivative equal to 4: \( x^2 - 3x = 4 \).
4Step 4: Solve the resulting equation
Rearrange and solve: \( x^2 - 3x - 4 = 0 \). Factor the quadratic equation to \( (x-4)(x+1) = 0 \). Solve for \( x \), giving \( x = 4 \) and \( x = -1 \).
5Step 5: Find the corresponding y-values
Substitute the \( x \)-values into \( g(x) \) to find the corresponding \( y \)-values. For \( x = 4 \), \( g(4) = \frac{1}{3}(4)^3 - \frac{3}{2}(4)^2 + 1 = \frac{64}{3} - 24 + 1 = \frac{5}{3} \). For \( x = -1 \), \( g(-1) = \frac{1}{3}(-1)^3 - \frac{3}{2}(-1)^2 + 1 = -\frac{1}{3} - \frac{3}{2} + 1 = -\frac{5}{6} \).
6Step 6: State the solution
The points where the tangent lines are parallel to the line \( 8x - 2y = 1 \) are \( (4, \frac{5}{3}) \) and \( (-1, -\frac{5}{6}) \).
Key Concepts
DifferentiationTangent LineQuadratic Equation
Differentiation
Differentiation is the process of finding the derivative of a function. A derivative represents the rate at which a function is changing at any given point. This is crucial for calculus, as it allows us to understand how functions behave and how their shapes change. In our specific problem, we deal with the function \(g(x) = \frac{1}{3} x^3 - \frac{3}{2} x^2 + 1\). To find the derivatives, apply the power rule of differentiation, which states that the derivative of \(x^n\) is \(n \cdot x^{n-1}\). This yields the derivative \(g'(x) = x^2 - 3x\).
By differentiating, we obtain a new function that represents the slopes of the tangent lines to the graph of the original function at various points. Differentiation functions as a tool for finding these instantaneous rates of change, which in our context, helps us find where these slopes (also called gradients) match another line's slope.
By differentiating, we obtain a new function that represents the slopes of the tangent lines to the graph of the original function at various points. Differentiation functions as a tool for finding these instantaneous rates of change, which in our context, helps us find where these slopes (also called gradients) match another line's slope.
Tangent Line
A tangent line to a curve at a given point is a straight line that "touches" the curve at that point. It has the same slope as the curve does at that precise point, but doesn't necessarily intersect the curve anywhere else. Finding the tangent line involves calculating the slope of the curve at the point of tangency.
In this exercise, we are specifically looking for points on the curve \(g(x)=\frac{1}{3}x^3-\frac{3}{2}x^2+1\) where the tangent line is parallel to the line described by the equation \(8x - 2y = 1\).
Parallel lines share the same slope. Thus, by setting the derivative of the function \(g'(x) = x^2 - 3x\) equal to the slope of the given line (which is 4), we can determine where the tangent line matches the slope of the original line: \(x^2 - 3x = 4\). Solving this equation helps us find the specific \(x\)-values where the curve has tangent lines parallel to the line \(8x - 2y = 1\).
In this exercise, we are specifically looking for points on the curve \(g(x)=\frac{1}{3}x^3-\frac{3}{2}x^2+1\) where the tangent line is parallel to the line described by the equation \(8x - 2y = 1\).
Parallel lines share the same slope. Thus, by setting the derivative of the function \(g'(x) = x^2 - 3x\) equal to the slope of the given line (which is 4), we can determine where the tangent line matches the slope of the original line: \(x^2 - 3x = 4\). Solving this equation helps us find the specific \(x\)-values where the curve has tangent lines parallel to the line \(8x - 2y = 1\).
Quadratic Equation
Quadratic equations are polynomial equations of the second degree, generally formulated as \(ax^2 + bx + c = 0\). Solving these equations can be achieved by various methods such as factoring, using the quadratic formula, or completing the square.
In our solution, we solve the quadratic equation \(x^2 - 3x - 4 = 0\) to find the points where the derivative \(g'(x)\) is equal to 4. This equation corresponds to the scenarios where the curve and the line have parallel tangent lines. Factoring the quadratic results in \((x-4)(x+1) = 0\). Solving for \(x\), we get two potential solutions: \(x = 4\) and \(x = -1\).
Once these solutions are found, they are substituted back into the original function \(g(x)\) to find the associated \(y\)-values. Thus, the resulting points on the graph are \((4, \frac{5}{3})\) and \((-1, -\frac{5}{6})\), where the tangent lines align with the given line.
In our solution, we solve the quadratic equation \(x^2 - 3x - 4 = 0\) to find the points where the derivative \(g'(x)\) is equal to 4. This equation corresponds to the scenarios where the curve and the line have parallel tangent lines. Factoring the quadratic results in \((x-4)(x+1) = 0\). Solving for \(x\), we get two potential solutions: \(x = 4\) and \(x = -1\).
Once these solutions are found, they are substituted back into the original function \(g(x)\) to find the associated \(y\)-values. Thus, the resulting points on the graph are \((4, \frac{5}{3})\) and \((-1, -\frac{5}{6})\), where the tangent lines align with the given line.
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