91Ó°ÊÓ

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.

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.

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.