Problem 13
Question
Use the method in Example 3 to find (a) the slope of the curve at the given point \(P,\) and \((\mathrm{b})\) an equation of the tangent line at \(P.\) \(y=x^{3}-12 x, \quad P(1,-11)\)
Step-by-Step Solution
Verified Answer
Slope: -9; Tangent line: \( y = -9x - 2 \).
1Step 1: Find the Derivative
To determine the slope of the curve at any point, we first need to find the derivative of the function. The given function is \( y = x^3 - 12x \). Using the power rule, the derivative \( y' \) can be calculated as: \[ y' = \frac{d}{dx}(x^3) - \frac{d}{dx}(12x) = 3x^2 - 12. \]
2Step 2: Evaluate the Derivative at P
The slope of the curve at point \( P(1, -11) \) is given by the derivative evaluated at \( x = 1 \). So, substituting \( x = 1 \) into \( y' \), we get:\[ y'(1) = 3(1)^2 - 12 = 3 - 12 = -9. \]Thus, the slope of the curve at \( P \) is \(-9\).
3Step 3: Use Point-Slope Form for Tangent Line
The equation of the tangent line can be found using the point-slope form of a line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1) = P(1, -11)\). Substituting the values, we get:\[ y - (-11) = -9(x - 1). \]
4Step 4: Simplify the Tangent Line Equation
Simplify the expression from the previous step to get the equation of the tangent line:\[ y + 11 = -9x + 9. \]Subtract 11 from both sides to find the equation of the tangent line:\[ y = -9x - 2. \]
Key Concepts
DerivativeSlope of a CurvePoint-Slope Form
Derivative
The concept of a derivative is foundational in calculus and crucial for understanding how a function behaves at any given point. Think of a derivative as a tool that tells us how a function changes as its input changes. This is incredibly useful for finding things like velocity, acceleration, and rates of change in various contexts.
Let's focus on how to find a derivative, specifically using power rule. If you have a function such as \( y = x^n \), the derivative, denoted as \( y' \) or \( \frac{dy}{dx} \), shows how \( y \) changes with respect to \( x \). Using the power rule, this derivative can be found by multiplying the power \( n \) by the coefficient of \( x \), and then reducing the power by one.
In the exercise, the function is \( y = x^3 - 12x \). By applying the power rule, we find:
Let's focus on how to find a derivative, specifically using power rule. If you have a function such as \( y = x^n \), the derivative, denoted as \( y' \) or \( \frac{dy}{dx} \), shows how \( y \) changes with respect to \( x \). Using the power rule, this derivative can be found by multiplying the power \( n \) by the coefficient of \( x \), and then reducing the power by one.
In the exercise, the function is \( y = x^3 - 12x \). By applying the power rule, we find:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -12x \) is \(-12 \).
Slope of a Curve
The slope of a curve at a given point is a measure of how steep the curve is at that point. In simple terms, when you have a slope, it tells you how much \( y \) changes for a change in \( x \). For curves, this slope can differ at each point along the curve.
To find the exact slope of a curve at a specific point, you'll need to evaluate the derivative at that point. In the given problem, we're interested in the point \( P(1, -11) \). To determine the slope at this point, you simply plug \( x = 1 \) into the derived formula \( y' = 3x^2 - 12 \).
This calculation yields \( y'(1) = 3(1)^2 - 12 = -9 \). Therefore, the slope of the curve at point \( P \) is \(-9\). This negative value indicates the curve is decreasing at this point, meaning it's slanting downwards as you move right.
To find the exact slope of a curve at a specific point, you'll need to evaluate the derivative at that point. In the given problem, we're interested in the point \( P(1, -11) \). To determine the slope at this point, you simply plug \( x = 1 \) into the derived formula \( y' = 3x^2 - 12 \).
This calculation yields \( y'(1) = 3(1)^2 - 12 = -9 \). Therefore, the slope of the curve at point \( P \) is \(-9\). This negative value indicates the curve is decreasing at this point, meaning it's slanting downwards as you move right.
Point-Slope Form
The point-slope form of a linear equation is vital when you intend to write the equation of a line that is tangent to a curve at a particular point. This method is especially useful when you know a point on the line and its slope.
The point-slope form is given by the equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point on the line. For example, if you have a point \( (1, -11) \) and the slope \(-9\), you can substitute these values into the formula to get:
The point-slope form is given by the equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point on the line. For example, if you have a point \( (1, -11) \) and the slope \(-9\), you can substitute these values into the formula to get:
- Start with \( y + 11 = -9(x - 1) \).
By expanding this, first distribute the \(-9\) to get: \(-9x + 9 \). - Simplify to obtain the line equation: \( y = -9x - 2 \).
Notice how this provides a linear equation that perfectly represents the tangent line at the point \( P \).
Other exercises in this chapter
Problem 12
Find the limits in Exercises \(11-22\) $$\lim _{x \rightarrow 2}\left(-x^{2}+5 x-2\right)$$
View solution Problem 13
At what points are the functions in Exercises \(13-30\) continuous? $$ y=\frac{1}{x-2}-3 x $$
View solution Problem 13
In Exercises \(13-22,\) find the limit of each rational function (a) as \(x \rightarrow \infty\) and \((b)\) as \(x \rightarrow-\infty\) . $$f(x)=\frac{2 x+3}{5
View solution Problem 13
Find the limits in Exercises \(11-18\) $$\lim _{x \rightarrow-2^{+}}\left(\frac{x}{x+1}\right)\left(\frac{2 x+5}{x^{2}+x}\right)$$
View solution