Chapter 3: Problem 51
Find all points on the graph of \(y=x^{3}-x^{2}\) where the tangent line is horizontal.
Short Answer
Expert verified
The points are \((0, 0)\) and \(\left(\frac{2}{3}, -\frac{4}{27}\right)\)."
Step by step solution
01
Understand when the tangent line is horizontal
The tangent line to a curve is horizontal when the slope of the tangent line is zero. The slope of the tangent is given by the derivative of the function. So, first, we need to find the derivative of the function.
02
Find the derivative of the function
Given the function is \[ y = x^3 - x^2 \]The derivative of this function with respect to \( x \) is found using the power rule. Thus, the derivative \( y' \) is:\[ y' = \frac{d}{dx}(x^3 - x^2) = 3x^2 - 2x \]
03
Set the derivative equal to zero
To find where the tangent line is horizontal, set the derivative equal to zero:\[ 3x^2 - 2x = 0 \]
04
Solve the equation
Factor out the common term \( x \) from the equation:\[ x(3x - 2) = 0 \]Set each factor to zero and solve for \( x \):\[ x = 0 \] or \[ 3x - 2 = 0 \]Solving \( 3x - 2 = 0 \), we get:\[ 3x = 2 \quad \Rightarrow \quad x = \frac{2}{3} \]
05
Find the y-coordinates for these x-values
Substitute the \( x \) values back into the original function to find the \( y \)-coordinates:For \( x = 0 \):\[ y = (0)^3 - (0)^2 = 0 \]For \( x = \frac{2}{3} \):\[ y = \left( \frac{2}{3} \right)^3 - \left( \frac{2}{3} \right)^2 \]Calculate each part:\[ \left( \frac{2}{3} \right)^3 = \frac{8}{27} \quad \text{and} \quad \left( \frac{2}{3} \right)^2 = \frac{4}{9} = \frac{12}{27} \]Substitute these into the equation:\[ y = \frac{8}{27} - \frac{12}{27} = -\frac{4}{27} \]
06
Identify the points with horizontal tangents
The points where the tangent line is horizontal are \((0, 0)\) and \(\left(\frac{2}{3}, -\frac{4}{27}\right)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives are a fundamental concept in calculus that describe how a function changes at any given point. They represent the slope of the tangent line to the curve of the function. To find the derivative of a function, especially a polynomial like our example, we often use the power rule. The power rule states that if you have a term of the form \( x^n \), its derivative is \( nx^{n-1} \). This makes it straightforward to calculate derivatives of polynomial functions. In our problem, given the function \( y = x^3 - x^2 \), the derivative was calculated using the power rule to be \( y' = 3x^2 - 2x \). This derivative function \( y' \) tells us the slope of the tangent line to the curve \( y = x^3 - x^2 \) at any point \( x \). Finding the derivative is the first crucial step in understanding the behavior of a function and is essential for resolving problems involving tangent lines.
Tangent Lines
Tangent lines are lines that touch a curve at exactly one point, giving us an instantaneous rate of change for the curve at that point. For polynomials and many other types of functions, tangent lines provide local linear approximations. In calculus, the slope of a tangent line at a particular point is given by the derivative of the function at that point. A horizontal tangent line implies a slope of zero, which means the derivative at that point is zero. For our function \( y = x^3 - x^2 \), to find where the tangent lines are horizontal, we set the derivative equal to zero:
- First, find the derivative: \( y' = 3x^2 - 2x \).
- Set \( y' = 0 \) to find the critical points: \( 3x^2 - 2x = 0 \).
- Factoring gives us: \( x(3x - 2) = 0 \), leading to two solutions: \( x = 0 \) and \( x = \frac{2}{3} \).
Graph Analysis
Graph analysis involves studying the behavior of a graph of a function to understand various properties, such as where the function increases, decreases, or has constant behavior.In the context of this problem, graph analysis includes identifying points on the graph where the tangent line is horizontal, which corresponds to the derivative being zero. We found two such points: \((0, 0)\) and \( \left( \frac{2}{3}, -\frac{4}{27} \right) \). These points are vital because:
- They indicate where the function momentarily stops increasing or decreasing, breaking the monotony.
- In some contexts, these points can be associated with local maxima, minima, or saddle points, depending on further analysis of the second derivative and the surrounding behavior of the function.