91Ó°ÊÓ

When graphing polynomial functions, it's important to understand the basic shape and characteristics of the function. For the cubic function provided in the exercise, the equation is \(f(x) = x^3 - 147x + 286\). This is a standard form of a cubic polynomial, characterized by three main components: the leading term \(x^3\), the middle term \(-147x\), and the constant \(286\).

With a positive leading coefficient (in this case, 1), the graph generally starts in the third quadrant and ends in the first quadrant, forming an elongated 'S' shape. This gives us our initial understanding of the graph's direction and general shape.

When creating the graph, one crucial step involves determining where the graph intersects the axes. The x-intercepts are found by solving the equation \(f(x) = 0\), and they represent the points where the graph crosses the x-axis. These can often be approximated using numerical methods or graphing technology. Meanwhile, the y-intercept, where the graph crosses the y-axis, is found by evaluating \(f(0)\), giving the point \((0, 286)\).

Overall, understanding these initial graphical features provides a good foundation before diving into more complex characteristics of the function.

Critical points in calculus are essential for identifying where a function's graph changes direction. These are the points where the derivative of the function is either zero or undefined. In our cubic function \(f(x) = x^3 - 147x + 286\), we calculate the first derivative to find these points, which is \(f'(x) = 3x^2 - 147\).

By setting the first derivative equal to zero, we solve for the critical points: \(3x^2 - 147 = 0\), leading to \(x^2 = 49\) and \(x = \pm 7\).

• At \(x = -7\), we find a critical point leading to a potential relative maximum.
• At \(x = 7\), the critical point indicates a potential relative minimum.

To confirm whether these points are maxima or minima, one can either evaluate the function at these points or use the second derivative test. In this case, evaluating \(f(-7)\) and \(f(7)\) provides the respective y-values for these critical points.

Therefore, the function has a relative maximum at \((-7, 980)\) and a relative minimum at \((7, -388)\). Understanding these critical points helps in sketching a more accurate and detailed graph.

Understanding the intervals where a function is increasing or decreasing is crucial for correctly sketching its graph. For the cubic function in our exercise, the sign of the first derivative \(f'(x) = 3x^2 - 147\) informs us about these intervals.

These are determined by testing the intervals defined by the critical points \((-\infty, -7)\), \((-7, 7)\), and \((7, \infty)\):

• For \(x < -7\), we have \(f'(x) > 0\), indicating the function is increasing in this interval.
• For \(-7 < x < 7\), \(f'(x) < 0\), so the function decreases.
• Lastly, for \(x > 7\), \(f'(x) > 0\), and the function is again increasing.

These tests allow us to build a comprehensive view of the function's behavior over its entire domain. Knowing where a function increases or decreases helps in understanding its overall shape and identifying the placement of the critical points effectively.

Incorporating these increasing and decreasing intervals into the graph provides a richer understanding of the function's behavior, ensuring the graph accurately reflects the given function.