Problem 6
Question
\(5-8\) Find an equation of the tangent line to the curve at the given point. $$y=2 x^{3}-5 x,(-1,3)$$
Step-by-Step Solution
Verified Answer
The equation of the tangent line is \(y = x + 4\).
1Step 1: Differentiate the Function
First, we need to find the derivative of the given function to determine the slope of the tangent line at the given point. The function is \(y = 2x^3 - 5x\). Differentiate this function with respect to \(x\):\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3 - 5x) = 6x^2 - 5 \]
2Step 2: Calculate the Slope at the Given Point
Next, substitute \(x = -1\) into the derivative to find the slope of the tangent line at the point \((-1, 3)\):\[ 6(-1)^2 - 5 = 6 \times 1 - 5 = 6 - 5 = 1 \]The slope of the tangent line at \((-1, 3)\) is 1.
3Step 3: Use the Point-Slope Form of the Equation of a Line
Now that we have the slope, use the point-slope formula for the equation of a line: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is \((-1, 3)\), and \(m = 1\):\[ y - 3 = 1(x + 1) \]
4Step 4: Simplify the Equation
Finally, simplify the equation from Step 3 to get the equation of the tangent line:Start with:\[ y - 3 = 1(x + 1) \]Simplify:\[ y - 3 = x + 1 \]Add 3 to both sides:\[ y = x + 4 \]
Key Concepts
Understanding DerivativesExploring the Point-Slope FormUnderstanding the Slope of a Tangent LineApplying Differentiation
Understanding Derivatives
Derivatives are fundamental in calculus. They represent the rate at which something changes. Imagine driving a car; the derivative is like your speedometer showing how fast you're going. In mathematics, the derivative of a function at a specific point gives the slope of the tangent line to the curve at that point.
For example, with the function \( y = 2x^3 - 5x \), finding its derivative means calculating \( \frac{dy}{dx} \). This shows how \( y \) changes with respect to changes in \( x \).
The derivative here is \( 6x^2 - 5 \). This expression tells us the slope of the tangent line for any value of \( x \). To find the exact slope at a specific point, we substitute the \( x \) value of that point into the derivative equation.
For example, with the function \( y = 2x^3 - 5x \), finding its derivative means calculating \( \frac{dy}{dx} \). This shows how \( y \) changes with respect to changes in \( x \).
The derivative here is \( 6x^2 - 5 \). This expression tells us the slope of the tangent line for any value of \( x \). To find the exact slope at a specific point, we substitute the \( x \) value of that point into the derivative equation.
Exploring the Point-Slope Form
The point-slope form is a handy tool for writing the equation of a line. This form is expressed as \( y - y_1 = m(x - x_1) \).
This form simplifies the task of finding the line equation next to a specific point with ease.
- \( y_1 \) and \( x_1 \) are the coordinates of a given point on the line.
- \( m \) represents the slope of the line.
This form simplifies the task of finding the line equation next to a specific point with ease.
Understanding the Slope of a Tangent Line
The slope of a tangent line is crucial when dealing with curves. This slope comes from the derivative of the function. Essentially, it tells us how steep the tangent is at a particular point.
In our exercise, by differentiating \( y = 2x^3 - 5x \) to get \( 6x^2 - 5 \), we determine the slope at \( x = -1 \) by substituting it into the derivative.
This yields a slope of \( 1 \). This slope means the tangent line rises 1 unit for every unit it moves right at the point \((-1,3)\). A slope of 1 describes a line at a 45-degree angle to the x-axis, which is neither steep nor flat.
Knowing this slope helps us paint the perfect picture of how the tangent line behaves and aligns with the curve at the specified point.
In our exercise, by differentiating \( y = 2x^3 - 5x \) to get \( 6x^2 - 5 \), we determine the slope at \( x = -1 \) by substituting it into the derivative.
This yields a slope of \( 1 \). This slope means the tangent line rises 1 unit for every unit it moves right at the point \((-1,3)\). A slope of 1 describes a line at a 45-degree angle to the x-axis, which is neither steep nor flat.
Knowing this slope helps us paint the perfect picture of how the tangent line behaves and aligns with the curve at the specified point.
Applying Differentiation
Differentiation is the process used to find a derivative. This method helps to determine how a function's value changes as its input changes. The rules of differentiation, like the power rule, product rule, and chain rule, simplify this process.
By applying them to \( y = 2x^3 - 5x \), we followed the power rule to obtain \( 6x^2 - 5 \). This expression is the derivative, giving us the slope for any \( x \) value.
Differentiation turns complex calculations into simpler functions that offer insight, such as the steepness or behavior of a function at points of interest. Mastery of this technique is valuable and often necessary in calculus. It lets us solve real-world problems involving rates of change efficiently.
By applying them to \( y = 2x^3 - 5x \), we followed the power rule to obtain \( 6x^2 - 5 \). This expression is the derivative, giving us the slope for any \( x \) value.
Differentiation turns complex calculations into simpler functions that offer insight, such as the steepness or behavior of a function at points of interest. Mastery of this technique is valuable and often necessary in calculus. It lets us solve real-world problems involving rates of change efficiently.
Other exercises in this chapter
Problem 5
Sketch the graph of a function that is continuous everywhere except at \(x=3\) and is continuous from the left at \(3 .\)
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Evaluate the limit and justify each step by indicating the appropriate Limit Law(s). $$\lim _{t \rightarrow-1}\left(t^{2}+1\right)^{3}(t+3)^{5}$$
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\(5-10\) Sketch the graph of an example of a function \(f\) that satisfies all of the given conditions. $$ \begin{array}{l}{\lim _{x \rightarrow 0^{+}} f(x)=\in
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