Problem 3 - Calculus with Concepts in Calculus

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 \((b, f(b))\). $$ f(x)=x^{3}-6 x ; a=-2, b=0 $$

Step-by-Step Solution

Answer

The number \( c \) in the interval \((-2, 0)\) is approximately \(-1.15\).

Step 1: Determine the Slope of the Secant Line

To find the slope of the line joining \((a, f(a))\) and \((b, f(b))\), calculate \((f(b) - f(a)) / (b - a)\).\Firstly, calculate \ f(a) = (-2)^3 - 6(-2) = -8 + 12 = 4 \,\Next, calculate \ f(b) = (0)^3 - 6(0) = 0 \.So, the slope of the line is \\[ m = \frac{f(0) - f(-2)}{0 - (-2)} = \frac{0 - 4}{2} = -2. \]

Step 2: Find the Derivative of the Function

The slope of the tangent line to the curve at any point \(x\) is given by the derivative \(f'(x)\). Differentiate the function: \(f(x) = x^3 - 6x\) gives \(f'(x) = 3x^2 - 6\).

Step 3: Set the Derivative Equal to the Secant Slope

We need the slope of the tangent (derivative) to match the secant slope found in Step 1. Set \(f'(x) = -2\). This gives \[ 3x^2 - 6 = -2. \]

Step 4: Solve the Equation for x

First, add 6 to both sides: \(3x^2 = 4\). Then divide by 3: \(x^2 = \frac{4}{3}\). Finally, take the square root of both sides to find \\[ x = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}}. \]

Step 5: Identify Values in the Interval

The interval we are interested in is \((-2, 0)\). Calculate the numerical values: \-\sqrt{\frac{4}{3}} = -\frac{2}{\sqrt{3}} \approx -1.15 \.Since \(-1.15\) is within \((-2, 0)\), it's a valid solution; \(\sqrt{\frac{4}{3}}\) is positive and outside the interval.

Key Concepts

Calculus

Calculus is a branch of mathematics that allows us to study changes and motion through derivatives and integrals. In this context, the derivative is a fundamental concept, as it represents the rate of change of a function. This is particularly useful for understanding how functions behave across different points.

• A derivative, denoted as \(f'(x)\), gives us the slope of a function at any given point.
• The process of finding a derivative is called differentiation.
• Derivatives can show us where a function is increasing, decreasing, or finds any local extrema.

In problems like the one discussed, derivatives are used to identify critical points where the tangent line matches certain criteria, such as being parallel to another line. This is a classic application of calculus and helps us understand the shape and slope of a curve in more detail.

Secant Line

A secant line is a line that connects two points on the graph of a function. It helps us determine the average rate of change of the function between those two points.

• The slope of the secant line between points \((a, f(a))\) and \((b, f(b))\) is calculated as \(\frac{f(b) - f(a)}{b - a}\).
• This slope provides an average value that represents how the function slopes between these two points.

In our exercise, we find the secant line slope between the points \(a = -2\) and \(b = 0\). This slope is crucial, as we use it to locate points on the function's graph where a tangent line has the same slope, according to the Mean Value Theorem.

Tangent Line

A tangent line to a function at a point is a straight line that just "touches" the graph of the function at that point. It represents the instantaneous rate of change of the function at that position, similar to viewing speed at a single moment in time.

• To find the slope of the tangent line at a point \(x\), we take the derivative \(f'(x)\) of the function.
• For the line to be tangent, it must intersect the function at exactly one point without crossing it.

In this scenario, we set the derivative equal to the secant line's slope. This tells us that wherever \(f'(x) = -2\), the slope of the curve at \(x\) is the same as that of the secant line, indicating parallelism. Finding these points helps identify where the tangent line is aligned with the secant line, adhering to the Mean Value Theorem.