Inflection Points: F(x) = 3x^4 + 29x^3 - 15x^2 + 2

Nov 11, 2025 by Admin 51 views

Hey guys! Today, we're diving into a fun little calculus problem: finding the inflection points of the function f(x) = 3x^4 + 29x^3 - 15x^2 + 2. Buckle up, it's gonna be a smooth ride!

Understanding Inflection Points

Before we jump into the nitty-gritty, let's quickly recap what inflection points actually are. An inflection point is a point on a curve where the concavity changes. Imagine you're driving along a road; if you transition from driving on a road that curves to the left to one that curves to the right (or vice versa), that transition point is kind of like an inflection point. Mathematically, this means the second derivative of the function changes its sign at that point. So, our mission is to find where this happens for our given function.

Why are inflection points important, you ask? Well, they give us valuable information about the behavior of a function. They help us understand where the function is increasing or decreasing at an increasing rate (concave up) and where it's increasing or decreasing at a decreasing rate (concave down). This knowledge is super useful in various fields, from economics (understanding growth rates) to physics (analyzing motion). Plus, it's a great exercise for our calculus muscles!

Step 1: Find the First Derivative

Alright, let's get our hands dirty with some derivatives! The first thing we need to do is find the first derivative of f(x). Remember, the power rule is our best friend here. The power rule states that if f(x) = ax^n, then f'(x) = nax^(n-1). Applying this to our function, we get:

f(x) = 3x^4 + 29x^3 - 15x^2 + 2

f'(x) = 12x^3 + 87x^2 - 30x

Piece of cake, right? We just multiplied each term by its exponent and then reduced the exponent by one. Notice that the constant term (2) disappears because its derivative is zero. This first derivative tells us about the slope of the tangent line at any point on the original curve. It's useful for finding critical points (where the slope is zero), but we need the second derivative to find inflection points.

Step 2: Find the Second Derivative

Now, for the second derivative! We need to differentiate f'(x) to get f''(x). Again, we'll use the power rule:

f'(x) = 12x^3 + 87x^2 - 30x

f''(x) = 36x^2 + 174x - 30

Awesome! This second derivative, f''(x), is what we're really interested in. It tells us about the concavity of the original function. If f''(x) > 0, the function is concave up (like a smile), and if f''(x) < 0, the function is concave down (like a frown). Inflection points occur where f''(x) changes sign, so we need to find where f''(x) = 0.

Step 3: Set the Second Derivative Equal to Zero and Solve

To find potential inflection points, we need to solve the equation f''(x) = 0. So, we have:

36x^2 + 174x - 30 = 0

This is a quadratic equation. To make our lives easier, let's simplify it by dividing everything by 6:

6x^2 + 29x - 5 = 0

Now, we need to solve this quadratic equation for x. There are several ways to do this: factoring, completing the square, or using the quadratic formula. In this case, factoring might be a bit tricky, so let's use the quadratic formula. Remember the quadratic formula? For an equation ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 6, b = 29, and c = -5. Plugging these values into the quadratic formula, we get:

x = (-29 ± √(29^2 - 4 * 6 * -5)) / (2 * 6)

x = (-29 ± √(841 + 120)) / 12

x = (-29 ± √961) / 12

x = (-29 ± 31) / 12

So, we have two possible solutions:

x_1 = (-29 + 31) / 12 = 2 / 12 = 1/6

x_2 = (-29 - 31) / 12 = -60 / 12 = -5

These are our potential inflection points! But we're not done yet. We need to make sure the concavity actually changes at these points.

Step 4: Test Intervals Around Potential Inflection Points

To confirm that x = 1/6 and x = -5 are indeed inflection points, we need to test the sign of f''(x) in the intervals around these points. This will tell us whether the concavity changes.

Let's create a number line and mark our potential inflection points:

---(-5)---(1/6)---

We need to test the intervals: (-∞, -5), (-5, 1/6), and (1/6, ∞).

Interval (-∞, -5): Choose a test point, say x = -6. Plug it into f''(x): f''(-6) = 36(-6)^2 + 174(-6) - 30 = 36(36) - 1044 - 30 = 1296 - 1044 - 30 = 222 Since f''(-6) > 0, the function is concave up in this interval.
Interval (-5, 1/6): Choose a test point, say x = 0. Plug it into f''(x): f''(0) = 36(0)^2 + 174(0) - 30 = -30 Since f''(0) < 0, the function is concave down in this interval.
Interval (1/6, ∞): Choose a test point, say x = 1. Plug it into f''(x): f''(1) = 36(1)^2 + 174(1) - 30 = 36 + 174 - 30 = 180 Since f''(1) > 0, the function is concave up in this interval.

Notice that the sign of f''(x) changes at both x = -5 and x = 1/6. This confirms that they are indeed inflection points!

Step 5: Find the y-coordinates of the Inflection Points

To find the complete coordinates of the inflection points, we need to plug our x-values back into the original function f(x) to find the corresponding y-values.

For x = -5: f(-5) = 3(-5)^4 + 29(-5)^3 - 15(-5)^2 + 2 = 3(625) + 29(-125) - 15(25) + 2 = 1875 - 3625 - 375 + 2 = -2123 So, one inflection point is (-5, -2123).
For x = 1/6: f(1/6) = 3(1/6)^4 + 29(1/6)^3 - 15(1/6)^2 + 2 = 3(1/1296) + 29(1/216) - 15(1/36) + 2 = 1/432 + 29/216 - 15/36 + 2 To add these fractions, we need a common denominator, which is 432: f(1/6) = 1/432 + (29 * 2)/432 - (15 * 12)/432 + (2 * 432)/432 = (1 + 58 - 180 + 864) / 432 = 743 / 432 ≈ 1.718 So, the other inflection point is (1/6, 743/432).

Conclusion

Alright, we did it! We found the inflection points of the function f(x) = 3x^4 + 29x^3 - 15x^2 + 2. They are:

(-5, -2123)
(1/6, 743/432)

These points mark where the concavity of the function changes. Remember, finding inflection points involves finding the second derivative, setting it equal to zero, and then testing intervals around the potential inflection points to confirm that the concavity actually changes. Hope this was helpful and happy calculating!