Finding Max/Min Points & Sketching Curves: A Calculus Guide

Hey guys! Let's dive into some calculus fun and figure out how to find the maximum and minimum points of a curve, and then, how to sketch that curve. We're going to use the example of the equation: $y = x^3 - 6x^2 + 9x + 6$. Don't worry, it's not as scary as it sounds! This is a classic calculus problem, and once you get the hang of it, you'll be able to tackle similar problems with ease. This guide will walk you through the steps, making sure you understand every single step.

Step 1: Find the Derivative

Alright, first things first, we need to find the derivative of the function. The derivative tells us the slope of the curve at any given point. To find the derivative of $y = x^3 - 6x^2 + 9x + 6$, we apply the power rule. If you're rusty on the power rule, it's pretty simple: for a term like $ax^n$, the derivative is $nax^{n-1}$. Let's break it down:

• For $x^3$, the derivative is $3x^2$ (3 * 1x^(3-1)).
• For $-6x^2$, the derivative is $-12x$ (-6 * 2x^(2-1)).
• For $9x$, the derivative is $9$ (9 * 1x^(1-1), and x^0 is 1).
• The constant term, 6, has a derivative of 0.

So, the derivative of our function is $y' = 3x^2 - 12x + 9$. This is super important because it helps us find the critical points, which are the potential locations of our maximum and minimum values.

Step 2: Find the Critical Points

Critical points are where the derivative is equal to zero or undefined. In our case, the derivative, $y' = 3x^2 - 12x + 9$, is a polynomial, so it's defined everywhere. Therefore, we only need to find where the derivative equals zero. To do that, we set the derivative equal to zero and solve for x:

$3x^2 - 12x + 9 = 0$

We can simplify this by dividing the entire equation by 3:

$x^2 - 4x + 3 = 0$

Now we need to factor this quadratic equation. We're looking for two numbers that multiply to 3 and add up to -4. Those numbers are -1 and -3. So, the factored form of the equation is:

$(x - 1)(x - 3) = 0$

This gives us two solutions for x: $x = 1$ and $x = 3$. These are our critical points! These are the x-values where the slope of the curve is zero, meaning they could be maximums, minimums, or points of inflection.

Step 3: Determine if Critical Points are Maximums or Minimums

Now, how do we know if these critical points are maximums, minimums, or neither? There are a couple of ways to do this:

Method 1: The Second Derivative Test

This is a quick and easy method. We find the second derivative of the original function. The second derivative tells us about the concavity of the function (whether it curves upwards or downwards). To find the second derivative, we take the derivative of the first derivative:

• First derivative: $y' = 3x^2 - 12x + 9$
• Second derivative: $y'' = 6x - 12$

Now we plug in our critical points into the second derivative:

• For $x = 1$: $y'' = 6(1) - 12 = -6$. Since the second derivative is negative, the curve is concave down at $x = 1$, which means we have a local maximum.
• For $x = 3$: $y'' = 6(3) - 12 = 6$. Since the second derivative is positive, the curve is concave up at $x = 3$, which means we have a local minimum.

Method 2: The First Derivative Test

This method involves analyzing the sign of the first derivative around the critical points. This shows how the function is increasing or decreasing:

• Choose a value less than 1 (e.g., 0). Plug it into the first derivative $y' = 3x^2 - 12x + 9$. $y'(0) = 9$. The derivative is positive, so the function is increasing to the left of x = 1.
• Choose a value between 1 and 3 (e.g., 2). $y'(2) = -3$. The derivative is negative, so the function is decreasing between x = 1 and x = 3.
• Choose a value greater than 3 (e.g., 4). $y'(4) = 9$. The derivative is positive, so the function is increasing to the right of x = 3.

This tells us that we have a local maximum at $x = 1$ and a local minimum at $x = 3$, which confirms what we found with the second derivative test.

Step 4: Find the y-values of the Maximum and Minimum Points

To find the actual coordinates of the maximum and minimum points, we plug the x-values of our critical points back into the original equation $y = x^3 - 6x^2 + 9x + 6$:

• For $x = 1$: $y = (1)^3 - 6(1)^2 + 9(1) + 6 = 1 - 6 + 9 + 6 = 10$. So, the local maximum point is (1, 10).
• For $x = 3$: $y = (3)^3 - 6(3)^2 + 9(3) + 6 = 27 - 54 + 27 + 6 = 6$. So, the local minimum point is (3, 6).

Step 5: Sketching the Curve

Now we have enough information to sketch the curve. Here's a summary of what we know:

• Local maximum at (1, 10).
• Local minimum at (3, 6).

Also, a good thing to do is find the y-intercept. This is the point where the curve crosses the y-axis (where x = 0). Plug x = 0 into the original equation: $y = (0)^3 - 6(0)^2 + 9(0) + 6 = 6$. So, the y-intercept is (0, 6).

Now, let's sketch it. You can do this by hand or using graphing software. Here's the general process:

1. Plot the Key Points: Plot the maximum point (1, 10), the minimum point (3, 6), and the y-intercept (0, 6).
2. Consider the End Behavior: The leading term of our equation is $x^3$, which is positive. This tells us that as x goes to negative infinity, y goes to negative infinity, and as x goes to positive infinity, y goes to positive infinity.
3. Connect the Dots: Starting from the left, the curve goes up, passes through the y-intercept (0, 6), reaches the maximum at (1, 10), goes down, passes through the minimum at (3, 6), and then goes up again.
4. Make it Smooth: Draw a smooth curve that connects these points, keeping in mind the increasing/decreasing behavior we determined earlier.

You should end up with an 'S' shaped curve. The curve increases, reaching a local maximum, then decreases, reaching a local minimum, and then continues to increase again.

Step 6: Putting it All Together

• Original equation: $y = x^3 - 6x^2 + 9x + 6$
• Derivative: $y' = 3x^2 - 12x + 9$
• Critical points: $x = 1, x = 3$
• Second derivative: $y'' = 6x - 12$
• Local maximum: (1, 10)
• Local minimum: (3, 6)
• Y-intercept: (0, 6)

By following these steps, you've successfully found the maximum and minimum points and sketched the graph of the curve! It might seem like a lot at first, but with practice, it becomes second nature. Keep practicing, and you'll become a calculus pro in no time! Remember to always find the derivative, find the critical points, use the second derivative or first derivative test to classify them, and plug them back into the original equation to find the y-values. And finally, sketch your curve!

I hope this guide has been helpful, guys! Keep up the great work and always remember to have fun with math! Happy calculating!