Problem 7 - 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 \((b, f(b))\). $$ f(x)=x^{3}-3 x^{2}+3 x+1 ; a=-2, b=2 $$

Step-by-Step Solution

Answer

The value of \(c\) is \(1 - \frac{\sqrt{21}}{3}\).

Step 1: Understand the Problem

To solve the problem, we need to find the values of \(c\) in the interval \((-2, 2)\) where the tangent to \(f(x)=x^3-3x^2+3x+1\) is parallel to the line joining \((-2, f(-2))\) and \((2, f(2))\).

Step 2: Calculate \(f(a)\) and \(f(b)\)

First, evaluate the function at the endpoints of the interval.\ \(f(-2)=(-2)^3-3(-2)^2+3(-2)+1 = -8 - 12 - 6 + 1 = -25\).\ \(f(2) = 2^3 - 3(2)^2 + 3(2) + 1 = 8 - 12 + 6 + 1 = 3\).The points are \((-2, -25)\) and \((2, 3)\).

Step 3: Find the Slope of the Secant Line

Calculate the slope of the line connecting \((-2, -25)\) and \((2, 3)\). The formula for the slope \(m\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).So, \(m = \frac{3 - (-25)}{2 - (-2)} = \frac{3 + 25}{2 + 2} = \frac{28}{4} = 7\).

Step 4: Find the Derivative of \(f(x)\)

To find where the tangent to \(f(x)\) is parallel to the secant line, find \(f'(x)\) and solve \(f'(x) = 7\).The derivative is \(f'(x) = 3x^2 - 6x + 3\).

Step 5: Solve \(f'(x) = 7\)

We need to solve the equation \(3x^2 - 6x + 3 = 7\).Simplify to get \(3x^2 - 6x + 3 - 7 = 0\) or \(3x^2 - 6x - 4 = 0\).

Step 6: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve \(3x^2 - 6x - 4 = 0\), with \(a = 3, b = -6, c = -4\).- Discriminant \(\Delta = (-6)^2 - 4 \times 3 \times (-4) = 36 + 48 = 84\).- Roots are \(x = \frac{6 \pm \sqrt{84}}{6} = \frac{6 \pm 2\sqrt{21}}{6} = 1 \pm \frac{\sqrt{21}}{3}\).

Step 7: Verify Roots in the Interval

Find the \(c\) values in the interval \((-2, 2)\):\(1 + \frac{\sqrt{21}}{3} \approx 2.5275\) which is outside the interval.\(1 - \frac{\sqrt{21}}{3} \approx -0.5275\) is within the interval \((-2, 2)\).Hence, the only valid \(c\) is \(1 - \frac{\sqrt{21}}{3}\).

Key Concepts

Function Derivatives

The derivative of a function is a key concept in calculus. It represents the rate of change or the slope of the function at any given point. For the function \( f(x) = x^3 - 3x^2 + 3x + 1 \), the derivative \( f'(x) \) tells us how the function's value changes as \( x \) changes.
To find \( f'(x) \), you apply the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). So, for each term in the function:

• \( x^3 \) becomes \( 3x^2 \).
• \(-3x^2 \) becomes \(-6x \).
• \(3x\) becomes \(3\).
• The constant \(1\) becomes \(0\).

Putting it all together, the derivative of \( f(x) \) is \( f'(x) = 3x^2 - 6x + 3 \).
This derivative will be used to find where the tangent line to the curve is parallel to a given line, which requires setting the derivative equal to the slope of that line.

Quadratic Equation Solutions

Solving a quadratic equation is a fundamental skill in algebra that often comes up in calculus problems. A quadratic equation typically takes the form \( ax^2 + bx + c = 0 \). In our example, the equation \( 3x^2 - 6x - 4 = 0 \) needs to be solved to find where the derivative equals the slope of the secant line.
To solve it, the quadratic formula is used, which is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a = 3 \), \( b = -6 \), and \( c = -4 \). The discriminant \( \Delta = b^2 - 4ac \) is calculated first, which helps to determine the nature of the roots.
For this quadratic, \( \Delta = 36 + 48 = 84 \), indicating two real and distinct roots exist. Substituting into the quadratic formula gives the roots \( x = \frac{6 \pm 2\sqrt{21}}{6} \), leading to the solutions \( x = 1 \pm \frac{\sqrt{21}}{3} \).
It is crucial to check whether these solutions fall within the specified interval.

Slope Calculation

Understanding slope calculation is crucial for this problem. The slope of a line measures its steepness and is found using the formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). This formula calculates the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\).
For the secant line joining the points \((-2, -25)\) and \((2, 3)\) on the graph of \( f(x) \), the slope is calculated as:

• Numerator: \( y_2 - y_1 = 3 - (-25) = 3 + 25 = 28 \).
• Denominator: \( x_2 - x_1 = 2 - (-2) = 2 + 2 = 4 \).

Thus, the slope \( m = \frac{28}{4} = 7 \).
This slope is critical as it is compared to the derivative to determine points where the tangent to the curve is parallel to this secant line. Therefore, solving \( f'(x) = 7 \) provides the desired values of \( c \) where the tangent is parallel to the secant.