91Ó°ÊÓ

To find the intervals of increase and decrease for a function, we first need to compute the first derivative. This helps us understand how the function behaves as the input values change. For the given function, the first derivative is calculated as follows:

Given function: \(f(x) = x^3 - 3x^2 + 2\)
First derivative: \(f'(x) = 3x^2 - 6x\)

Next, we find the critical points by setting the first derivative to zero:
\(3x^2 - 6x = 0\)
Solve for \(x\): \(x(3x - 6) = 0 \Rightarrow x = 0\) or \(x = 2\)

Now, we analyze the sign of \(f'(x)\) in the intervals determined by the critical points. These intervals are:
- \((-\infty, 0)\): Choose \(x = -1\), \(f'(-1) > 0\) (increasing)
- \((0, 2)\): Choose \(x = 1\), \(f'(1) < 0\) (decreasing)
- \((2, \infty)\): Choose \(x = 3\), \(f'(3) > 0\) (increasing)

• Increasing: \((-\infty, 0)\) and \((2, \infty)\)
• Decreasing: \((0, 2)\)

The first derivative \(f'(x)\) of a function helps us identify critical points and the intervals of increase and decrease. It is derived from the original function and gives us the slope of the tangent line at any point on the curve. In our function:

Original function: \(f(x) = x^3 - 3x^2 + 2\)
First derivative: \(f'(x) = 3x^2 - 6x\)

Finding the derivative is the first key step in analyzing the function because it helps us understand where the function is rising or falling. Setting the first derivative to zero helps pinpoint the exact critical points which partition the graph into segments.

Critical points are found by solving:
\(3x^2 - 6x = 0\)
This gives us critical points \(x = 0\) and \(x = 2\). These points divide the number line into different intervals where the sign of \(f'(x)\) will indicate whether the function is increasing or decreasing.

The second derivative of a function, denoted as \(f''(x)\), provides information on the concavity of the function. Concavity describes the direction the function curves. For the given function:

First derivative: \(f'(x) = 3x^2 - 6x\)
Second derivative: \(f''(x) = 6x - 6\)

Next, we set the second derivative to zero to find potential points of inflection (where the concavity changes):
\(6x - 6 = 0\)
Solve for \(x\): \(x = 1\)

Let's evaluate the concavity in the intervals split by the point of inflection:
- \((-\infty, 1)\): Choose \(x = 0\), \(f''(0) < 0\) (concave down)
- \((1, \infty)\): Choose \(x = 2\), \(f''(2) > 0\) (concave up)

The sign of the second derivative within these intervals tells us:

• Concave down: \((-\infty, 1)\)
• Concave up: \((1, \infty)\)

Critical points are where the first derivative is zero or undefined. These points can correspond to local maxima, minima, or saddle points in the function. To find the critical points for our function:

First derivative: \(f'(x) = 3x^2 - 6x\)
Set first derivative to zero:
\(3x^2 - 6x = 0\)
Solve for \(x\): \(x = 0\) or \(x = 2\)

These points are critical because they indicate where the rate of change of the function goes from positive to negative or vice versa. Evaluating the function at these critical points:

\(f(0) = 0^3 - 3(0)^2 + 2 = 2\)
\(f(2) = 2^3 - 3(2)^2 + 2 = -2\)

Thus, the critical points are \((0, 2)\) and \((2, -2)\). These points help in determining where the peaks and valleys of the graph are located.

Points of inflection occur where the function changes its concavity. These are found by setting the second derivative to zero and solving for \(x\). For our function:

Second derivative: \(f''(x) = 6x - 6\)

Set second derivative to zero:
\(6x - 6 = 0\)
Solve for \(x\): \(x = 1\)

Now, evaluate the original function at this point:

\(f(1) = 1^3 - 3(1)^2 + 2 = 0\)

So, the point of inflection is \((1, 0)\). This is where the function changes from concave down to concave up. Points of inflection are crucial because they indicate a change in the behavior of the function’s curve. They provide deeper insight into the function's shape and structure. By understanding these points, we can more accurately sketch the graph and understand how it bends and shifts.