Question

Find all points on the graph of \(y=\frac{1}{3} x^{3}+x^{2}-x\) where the tangent line has slope 1 .

Short answer

The points are \((-1 + \sqrt{3}, y_1)\) and \((-1 - \sqrt{3}, y_2)\).

Step-by-step solution

Recognize the Problem

We need to find points on the curve where the tangent line has a slope of 1. This means finding where the derivative of the function is equal to 1.

Find the Derivative

Differentiate the function with respect to \(x\). If \(y = \frac{1}{3}x^{3} + x^{2} - x\), then the derivative is \(y' = x^{2} + 2x - 1\).

Set the Derivative Equal to 1

Since we want the slope to be 1, set the derivative \(y' = x^{2} + 2x - 1\) equal to 1. So, solve: \(x^{2} + 2x - 1 = 1\).

Solve the Equation

Rearrange the equation to \(x^{2} + 2x - 2 = 0\). This is a quadratic equation we need to solve using the quadratic formula or by factoring.

Applying the Quadratic Formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\). Here, \(a = 1\), \(b = 2\), \(c = -2\). Substitute these into the formula:\[x = \frac{-2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}\ = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2}\ = -1 \pm \sqrt{3}\].

Calculate the Corresponding y-values

The solutions from the quadratic equation are \(x = -1 + \sqrt{3}\) and \(x = -1 - \sqrt{3}\). Substitute these \(x\)-values back into the original function to find the corresponding \(y\)-values.

Find y for each x-value

For \(x = -1 + \sqrt{3}\), substitute into the original equation:\[y = \frac{1}{3}(-1 + \sqrt{3})^3 + (-1 + \sqrt{3})^2 - (-1 + \sqrt{3})\].Simplify this to get the \(y\)-value.For \(x = -1 - \sqrt{3}\), substitute similarly:\[y = \frac{1}{3}(-1 - \sqrt{3})^3 + (-1 - \sqrt{3})^2 - (-1 - \sqrt{3})\].Simplify this to find the \(y\)-value.

Solution Points

After performing calculations, find the points as \((-1 + \sqrt{3}, y_1)\) and \((-1 - \sqrt{3}, y_2)\), where \(y_1\) and \(y_2\) are the specific values obtained from substitution and simplification. Therefore, these are the points on the graph with a tangent slope of 1.

Key concepts

Understanding Derivatives

Calculating the derivative of a function is like finding the slope of the tangent line at any point on a curve. This concept is essential in calculus. A derivative gives us crucial information about the rate at which a function changes.
The derivative of a function, often denoted as \( y' \) or \( f'(x) \), reflects how the \( y \)-value of a function changes as \( x \) changes. Essentially, it indicates the slope of the tangent line to the curve of the function at any particular point.
To find the derivative of a function, apply the standard rules of differentiation. For example, in our exercise:

• Function given is \( y = \frac{1}{3}x^3 + x^2 - x \).
• The derivative is computed to be \( y' = x^2 + 2x - 1 \).

Here, \( y' = x^2 + 2x - 1 \) tells us how steep the curve is at any point \( x \). Finding when this expression equals 1 gives us where the slope of the tangent is precisely 1.

Solving Quadratic Equations

A quadratic equation is any equation of the form \( ax^2 + bx + c = 0 \). It is fundamental in algebra and arises in various mathematical problems, including finding maximum or minimum values of a function.
In our exercise, after setting the derivative equal to 1, we obtained the quadratic equation \( x^2 + 2x - 2 = 0 \). There are multiple methods to solve a quadratic equation, such as:

• Factoring: Expressing the equation as a product of two binomials.
• Completing the square: Rewriting the equation in the form \( (x+p)^2 = q \).
• Quadratic formula: Using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Employing the quadratic formula is often straightforward, especially when factoring is complex or infeasible.
For our specific equation, solving yields \( x = -1 \pm \sqrt{3} \), providing the \( x \)-values where the tangent line's slope equals 1.

Using Graph Points to Solve Problems

Graph points, specifically \( (x, y) \) pairs, are essential for understanding the position and behavior of functions graphically. They tell us precisely where specific trends or values occur on a graph.
In the exercise, after calculating the \( x \)-values using the quadratic equation, we need to find the corresponding \( y \)-values of the original function \( y = \frac{1}{3}x^3 + x^2 - x \). This involves substituting each \( x \)-value back into the function:

• For \( x = -1 + \sqrt{3} \), substitute to find \( y_1 \).
• Similarly, for \( x = -1 - \sqrt{3} \), calculate to get \( y_2 \).

By doing so, we obtain complete points \((-1 + \sqrt{3}, y_1)\) and \((-1 - \sqrt{3}, y_2)\) on the graph. These points confirm where the tangent line has the desired slope, and help visualize the solution behavior on a graph.