91Ó°ÊÓ

Differentiation is a fundamental concept in calculus and is used to find the rate at which a function changes. When you "differentiate" a function, you're essentially finding its derivative. The derivative provides a way to calculate the slope of the tangent line to the curve at any given point. This becomes instrumental in understanding how the function behaves.

For the exercise, you were given the function \( y = x^3 - 2x^2 + x + 1 \). To differentiate this function:

• The derivative of \( x^3 \) is \( 3x^2 \).
• The derivative of \( -2x^2 \) is \( -4x \).
• The derivative of \( x \) is \( 1 \).
• And the derivative of a constant, \( 1 \), is \( 0 \).

So, we assemble these to form the derivative: \( y' = 3x^2 - 4x + 1 \). This derivative is crucial because it will help us locate where the horizontal tangents of the curve occur.

The quadratic formula is a handy tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). In cases where factoring is challenging or impossible, the quadratic formula comes in handy:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the exercise's solution, once the derivative \( 3x^2 - 4x + 1 \) was found, we equated it to zero: \( 3x^2 - 4x + 1 = 0 \). This gives us a quadratic equation that we can solve using the quadratic formula.

In our case, \( a = 3 \), \( b = -4 \), and \( c = 1 \). Plug these into the quadratic formula:

• Calculate the discriminant: \( b^2 - 4ac = (-4)^2 - 4 \times 3 \times 1 = 16 - 12 = 4 \)
• Then substitute into the formula: \( x = \frac{4 \pm \sqrt{4}}{6} \)

Simplifying this gives the solutions \( x = 1 \) and \( x = \frac{1}{3} \). These are the x-values where the curve has horizontal tangents.

Horizontal tangents occur on a curve when the derivative (or slope) of a function is zero. This means that at specific points, the curve doesn't rise or fall; instead, it has a flat tangent line. To find these points, we set the derivative equal to zero and solve for \( x \).In our exercise, the derivative of the function \( y = x^3 - 2x^2 + x + 1 \) is \( y' = 3x^2 - 4x + 1 \). Setting this equal to zero:

• \( 3x^2 - 4x + 1 = 0 \)

Using the quadratic formula, we discovered that \( x = 1 \) and \( x = \frac{1}{3} \) are where the horizontal tangents occur. These points indicate where the rate of change of the function is zero, providing insight into the function's behavior at these critical points.