Analyzing The Cubic Function: F(x) = 2x³ + 2x² + 3

Nov 28, 2025 by Admin 51 views

Hey guys! Let's dive deep into the analysis of a fascinating cubic function: f(x) = 2x³ + 2x² + 3. Understanding the behavior of functions is super important in calculus and beyond. This function, with its cubic term, offers a great opportunity to explore various calculus concepts. We'll break down everything from its domain and range to the nitty-gritty details of its concavity and inflection points. Buckle up, because we're about to embark on a mathematical adventure!

Domain and Range Unveiled

First things first: domain and range. The domain of a function refers to all possible input values (x-values) for which the function is defined, while the range refers to all possible output values (y-values) that the function can produce. For our cubic function, f(x) = 2x³ + 2x² + 3, the domain is all real numbers. This is because we can plug in any real number for 'x', and the function will produce a valid output. There are no restrictions like square roots of negative numbers or division by zero that would limit our input values. In interval notation, we express the domain as (-∞, ∞).

Now, let’s talk about the range. Because of the nature of cubic functions – they extend to both positive and negative infinity – the range of f(x) = 2x³ + 2x² + 3 is also all real numbers. As x goes towards positive infinity, so does f(x). As x goes toward negative infinity, f(x) also goes towards negative infinity. This behavior ensures that the function covers all possible y-values. So, the range is also expressed as (-∞, ∞). Pretty neat, right?

Uncovering Intercepts: Where the Function Crosses the Axes

Next, let's find the intercepts. These are the points where the function crosses the x-axis (x-intercepts) and the y-axis (y-intercept). Finding intercepts gives us key points to sketch the graph and provides valuable information about the function's behavior.

To find the y-intercept, we set x = 0 and solve for f(x). So, f(0) = 2(0)³ + 2(0)² + 3 = 3. Therefore, the y-intercept is at the point (0, 3). This tells us that the graph of the function crosses the y-axis at the point where y = 3.

Finding the x-intercepts is a bit trickier. We set f(x) = 0 and solve for x. This means we need to solve the equation 2x³ + 2x² + 3 = 0. Unfortunately, this cubic equation does not factor easily, and we can't find rational roots using simple algebraic methods. This is where numerical methods, like the Newton-Raphson method or a graphing calculator, come in handy. These tools help us approximate the real roots. In this case, there is one real root, which is approximately x ≈ -1.33. Hence, there is one x-intercept at approximately (-1.33, 0). Keep in mind that cubic functions can have up to three x-intercepts, but this one only has one real root, meaning the graph crosses the x-axis only once in the real number plane.

Critical Points and Intervals of Increase/Decrease: The Function's Rise and Fall

Understanding the critical points and intervals of increase/decrease is essential for sketching the graph and understanding the function's behavior. Critical points are the points where the derivative of the function is either equal to zero or undefined. These points often mark the locations of local maxima or minima.

To find these, we first need to find the derivative of f(x). Using the power rule, we get f'(x) = 6x² + 4x. Now, we set f'(x) = 0 and solve for x: 6x² + 4x = 0. We can factor this to get 2x(3x + 2) = 0, which gives us two critical points: x = 0 and x = -2/3. These are the x-coordinates of the critical points.

To determine whether these critical points are local maxima or minima, we can use the second derivative test. First, let's find the second derivative: f''(x) = 12x + 4. Now, we evaluate the second derivative at the critical points.

At x = 0, f''(0) = 4, which is positive. This means the function is concave up at x = 0, and we have a local minimum. To find the y-value, we plug x = 0 into the original function: f(0) = 3. So, we have a local minimum at the point (0, 3).
At x = -2/3, f''(-2/3) = -4, which is negative. This means the function is concave down at x = -2/3, and we have a local maximum. To find the y-value, we plug x = -2/3 into the original function: f(-2/3) = 2(-2/3)³ + 2(-2/3)² + 3 = 3.37. So, we have a local maximum at the point (-2/3, 3.37). Well done, we have the critical points!

To determine the intervals of increase and decrease, we analyze the sign of the first derivative, f'(x), on the intervals defined by the critical points. Remember, the function is increasing where f'(x) > 0 and decreasing where f'(x) < 0.

For x < -2/3: f'(-1) = 6(-1)² + 4(-1) = 2 > 0. The function is increasing.
For -2/3 < x < 0: f'(-1/3) = 6(-1/3)² + 4(-1/3) = -2/3 < 0. The function is decreasing.
For x > 0: f'(1) = 6(1)² + 4(1) = 10 > 0. The function is increasing. Therefore, the function increases in the intervals (-∞, -2/3) and (0, ∞), and decreases in the interval (-2/3, 0).

Unveiling Concavity and Points of Inflection

Now, let's look at concavity and points of inflection. Concavity describes the shape of the curve: whether it's curving upwards (concave up) or downwards (concave down). Points of inflection are the points where the concavity changes.

We already found the second derivative: f''(x) = 12x + 4. To find the points of inflection, we set f''(x) = 0 and solve for x: 12x + 4 = 0, which gives us x = -1/3. This is the x-coordinate of our potential inflection point.

To determine the intervals of concavity, we analyze the sign of the second derivative, f''(x), around the potential inflection point.

For x < -1/3: f''(-1) = -8 < 0. The function is concave down.
For x > -1/3: f''(0) = 4 > 0. The function is concave up.

This confirms that x = -1/3 is indeed a point of inflection. To find the y-coordinate, we plug x = -1/3 into the original function: f(-1/3) = 2(-1/3)³ + 2(-1/3)² + 3 = 3.18. Therefore, the point of inflection is (-1/3, 3.18). The function is concave down in the interval (-∞, -1/3) and concave up in the interval (-1/3, ∞). Understanding the concavity is very useful to understand the curve shape.

Graphical Representation

Finally, let's visualize all of this by describing what the graph looks like. The graph of f(x) = 2x³ + 2x² + 3 is a smooth, continuous curve that extends from negative infinity to positive infinity. It has a y-intercept at (0, 3) and one x-intercept near (-1.33, 0). The curve has a local maximum at approximately (-2/3, 3.37) and a local minimum at (0, 3). The concavity changes at the point of inflection (-1/3, 3.18), where the curve transitions from concave down to concave up. The graph will rise from negative infinity, reach its local maximum, decrease to its local minimum, and then rise again towards positive infinity. It looks like a slightly stretched