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When working with calculus, one important concept is the derivative, which represents the rate of change of a function with respect to one of its variables. In simpler terms, derivatives help us understand how fast something is changing at any given point. A common application of derivatives is to find the slope of the tangent line at a point on a curve. The derivative of a function is a crucial tool used to determine where tangent lines are horizontal. Horizontal tangent lines occur when the derivative is zero because this corresponds to a zero slope. This is why, in the given exercise, we start by finding the derivative of the function involved. To calculate the derivative of a function, various rules can be applied depending on the function's structure. In this exercise, we use the quotient rule because the function is expressed as a quotient.

The quotient rule is a technique used in calculus for finding the derivative of a quotient of two functions. It's essential when the function you need to differentiate is composed of a numerator and a denominator. The rule is structured to handle these calculations without violating the properties of derivatives.The quotient rule states that if you have a function given by \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative \( \frac{dy}{dx} \) is found using the formula:\[ \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \]This rule helps to differentiate complex fractional functions by keeping the operations organized. Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. In our exercise, \( u = x^2 + 1 \) and \( v = x - 1 \), leading to the derivative calculated to determine where the tangent line is horizontal.

The quadratic formula is a powerful method used to find solutions to quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula provides a way to find the roots of a quadratic equation, which could be significant values such as where a function's derivative equals zero.The quadratic formula is stated as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this formula, \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation. The term under the square root, \( b^2 - 4ac \), is known as the discriminant and determines the nature of the roots.For the given exercise, after simplifying the derivative, a quadratic equation is arrived at: \( x^2 - 2x - 1 = 0 \). We use the quadratic formula here to find \( x = 1 \pm \sqrt{2} \) as solutions, indicating the points where the derivative is zero, and thus the tangent line is horizontal.