Problem 63
Question
For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=x^{3}-2 $$
Step-by-Step Solution
Verified Answer
The tangent line is horizontal at the point (0, -2).
1Step 1: Understanding the Problem
We are given the function \( y = x^3 - 2 \) and need to find where the tangent line to the graph is horizontal. A tangent line is horizontal where its slope is zero.
2Step 2: Derive the Function
To find the slope of the tangent line, first compute the derivative of the function. Differentiate \( y = x^3 - 2 \) with respect to \( x \). \[ \frac{dy}{dx} = 3x^2 \]
3Step 3: Set the Derivative to Zero
Horizontal tangents occur where the derivative equals zero. Set the derivative \( 3x^2 \) equal to zero and solve for \( x \).\[ 3x^2 = 0 \]
4Step 4: Solve for \( x \)
Solve \( 3x^2 = 0 \): - Divide both sides by 3 to get \( x^2 = 0 \).- Take the square root of both sides to find \( x = 0 \).
5Step 5: Find the Corresponding \( y \)-value
To find the point on the graph, substitute \( x = 0 \) back into the original equation \( y = x^3 - 2 \) to get the corresponding \( y \)-value. \[ y = 0^3 - 2 = -2 \]
6Step 6: State the Result
The point on the graph where the tangent is horizontal is \((0, -2)\).
Key Concepts
Slope of a Tangent LineDerivative of a FunctionSolving Equations
Slope of a Tangent Line
The whole idea behind the tangent line is its relationship to a curve. A tangent line touches a curve at one unique point, and the slope of this line at that point represents the instantaneous rate of change of the function. In simpler terms, it's like peeking into the curve's behavior just at that specific point.
- A horizontal tangent line means that the slope at that point equals zero. Think of horizontal lines that run flat, neither rising nor falling.
- If the slope is zero, the curve right there is flattening out briefly, if just for an instant.
Derivative of a Function
Calculating the derivative of a function is like finding a special formula that tells us the slope of the tangent line at any given point on the curve. This formula changes depending on the point we're interested in.
- For the function in our example, the derivative is found using a straightforward method of differentiation. Given the function \( y = x^3 - 2 \), we differentiate with respect to \( x \).
- The power rule, one of the basic rules of differentiation, gives us \( \frac{dy}{dx} = 3x^2 \). It's really that simple. Just multiply by the power and reduce the power by one.
- This derivative, \( 3x^2 \), is our magical formula that reveals the slope of the tangent line for any \( x \).
Solving Equations
Once we've got the derivative, finding horizontal tangent lines boils down to a bit of simple equation solving.
- Set the derivative equal to zero because, remember, a horizontal line has a slope of zero.
- In our exercise, the derivative \( 3x^2 = 0 \) tells us exactly where the slope is zero. Solving this equation is straightforward; first, divide by 3 (which doesn't change anything significant) to get \( x^2 = 0 \).
- Next, take the square root of both sides, finding \( x = 0 \). What you've done is find the specific point or points where our curve flattens.
Other exercises in this chapter
Problem 62
Consider $$ f(x)=\frac{x^{2}}{(1+x)^{5}} $$ a) Find \(f^{\prime}(x)\) using the Quotient Rule and the Extended Power Rule. b) Note that \(f(x)=x^{2}(1+x)^{-5}\)
View solution Problem 62
Is the function given by \(f(x)=\left\\{\begin{array}{ll}\frac{x^{2}-4 x-5}{x-5}, & \text { for } x
View solution Problem 63
Find \(y^{\prime \prime}\) for each function. $$ y=\frac{\sqrt{x}+1}{\sqrt{x}-1} $$
View solution Problem 63
Graph each function and then find the specified limits. When necessary, state that the limit does not exist. $$ \begin{array}{l} g(x)=\left\\{\begin{array}{ll}
View solution