Problem 6
Question
Find all numbers \(c\) in the interval \((a, b)\) for which the line tangent to the graph of \(f\) is parallel to the line joining \((a, f(a))\) and \((b, f(b))\). $$ f(x)=x^{3}-2 ; a=-3, b=3 $$
Step-by-Step Solution
Verified Answer
All numbers \( c \) in \((-3, 3)\) such that \(f'(c) = 9\).
1Step 1: Determine the Slope of the Secant Line
First, calculate the slope of the line that joins the points \((a, f(a))\) and \((b, f(b))\). The formula for the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).Substituting the given points, we have:\[ f(a) = (-3)^3 - 2 = -29 \]\[ f(b) = (3)^3 - 2 = 25 \]\[ \text{Slope} = \frac{f(b) - f(a)}{b - a} = \frac{25 - (-29)}{3 - (-3)} = \frac{54}{6} = 9 \].
2Step 2: Identify the relevant trigonometric identities
Based on the given expression or equation, identify which trigonometric identities (Pythagorean, double-angle, sum/difference, etc.) are applicable.
3Step 3: Apply the identities and simplify
Apply the identified identities to transform the expression. Simplify step by step, combining like terms and reducing fractions where possible.
4Step 4: Solve or evaluate
If solving an equation, isolate the trigonometric function and find the angle(s). If evaluating, compute the final numerical value.
5Step 5: State the result
Express the final answer, including all solutions in the required domain if solving an equation.
6Step 6: Conclude with the answer
All numbers \( c \) in \((-3, 3)\) such that \(f'(c) = 9\).
Key Concepts
Tangent LineSecant LineDifferentiable Function
Tangent Line
A tangent line is a straight line that touches a curve at a single point. It's important because it represents the slope of the curve at that specific point, which is the same as the derivative of the function at that point. Understanding tangent lines helps us visualize how a function is changing at any given moment along the curve.
In our exercise, the tangent line we're looking for should be parallel to the secant line between the points \(a, f(a)\) and \(b, f(b)\). For two lines to be parallel, they must have the same slope. Therefore, the slope of the tangent line at some point \(c\) in the interval must equal 9, which is the slope of our secant line.
This idea links directly to the Mean Value Theorem, which states that if a function is continuous on a closed interval and differentiable on an open interval, there exists at least one point where the slope of the tangent (derivative) is equal to the slope of the secant line.
In our exercise, the tangent line we're looking for should be parallel to the secant line between the points \(a, f(a)\) and \(b, f(b)\). For two lines to be parallel, they must have the same slope. Therefore, the slope of the tangent line at some point \(c\) in the interval must equal 9, which is the slope of our secant line.
This idea links directly to the Mean Value Theorem, which states that if a function is continuous on a closed interval and differentiable on an open interval, there exists at least one point where the slope of the tangent (derivative) is equal to the slope of the secant line.
Secant Line
A secant line intersects a curve at two points, reminiscent of drawing a chord on a circle. The secant line gives us the average rate of change between those two points. It acts as a bridge or approximation of the curve between the points.
In our example, the secant line connects the points \(a, f(a)\) and \(b, f(b)\), which are \( (-3, -29) \) and \( (3, 25) \) respectively. Calculating the secant line's slope involves a simple application of the slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
In our example, the secant line connects the points \(a, f(a)\) and \(b, f(b)\), which are \( (-3, -29) \) and \( (3, 25) \) respectively. Calculating the secant line's slope involves a simple application of the slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
- The change in y-values is 25 - (-29) = 54.
- The change in x-values is 3 - (-3) = 6.
Differentiable Function
A differentiable function is one that has a derivative at each point in its domain. This means the function is smooth and not "broken" anywhere; there are no sharp corners or discontinuities.
In mathematical terms, a function \( f \) is differentiable on an interval if it is possible to compute the derivative \( f'(x) \) on every point of that interval. Differentiability guarantees that the function's behavior is predictable, which is crucial for finding tangent lines, as opposed to secant lines, which do not require differentiability.
The given function, \( f(x) = x^3 - 2 \, \) is a polynomial function, which is continuously differentiable everywhere in its domain. This attribute makes it eligible for the Mean Value Theorem, thus allowing us to find the point \( c \) where the tangent line has the same slope as the secant line.
The differentiability of the function guarantees that the tangent line at some point within the interval will have enough smoothness to exactly match the slope we calculated for the secant line, which is why we can identify such a point \( c \).
In mathematical terms, a function \( f \) is differentiable on an interval if it is possible to compute the derivative \( f'(x) \) on every point of that interval. Differentiability guarantees that the function's behavior is predictable, which is crucial for finding tangent lines, as opposed to secant lines, which do not require differentiability.
The given function, \( f(x) = x^3 - 2 \, \) is a polynomial function, which is continuously differentiable everywhere in its domain. This attribute makes it eligible for the Mean Value Theorem, thus allowing us to find the point \( c \) where the tangent line has the same slope as the secant line.
The differentiability of the function guarantees that the tangent line at some point within the interval will have enough smoothness to exactly match the slope we calculated for the secant line, which is why we can identify such a point \( c \).
Other exercises in this chapter
Problem 6
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According to one model, the time rate \(R\) at which a tumor grows is given by $$ R=A x \ln \frac{B}{x} \text { for } 0
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Determine the values of \(c\) at which \(f^{\prime}\) changes from positive to negative, or from negative to positive. $$ f(t)=\frac{1}{\sqrt{t-t^{2}}} $$
View solution