Question

Concavity Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$f(x)=2 x^{4}+8 x^{3}+12 x^{2}-x-2$$

Short answer

Are there any inflection points? Answer: The function is concave up on the intervals \((-\infty, -1)\) and \((-1, \infty)\). There are no inflection points.

Step-by-step solution

Find the first derivative of the function

To find the first derivative, we apply the power rule to each term of the function: $$\frac{d}{dx}(2x^4) = 8x^3, \ \frac{d}{dx}(8x^3) = 24x^2, \ \frac{d}{dx}(12x^2) = 24x, \ \frac{d}{dx}(-x) = -1, \ \frac{d}{dx}(-2) = 0$$ Combine all the terms: $$f'(x) = 8x^3 + 24x^2 + 24x - 1$$

Find the second derivative of the function

We differentiate the first derivative to find the second derivative: $$\frac{d^2}{dx^2}(8x^3) = 24x^2, \ \frac{d^2}{dx^2}(24x^2) = 48x, \ \frac{d^2}{dx^2}(24x) = 24, \ \frac{d^2}{dx^2}(-1) = 0$$ Combine all the terms: $$f''(x) = 24x^2 + 48x + 24$$

Find the critical points of the second derivative

The critical points are where the second derivative is zero or undefined. The second derivative is a quadratic function, so it doesn't have any points of undefined. To find the zeros, set the second derivative equal to zero and solve for x: $$24x^2 + 48x + 24 = 0$$ Divide the equation by 24 to simplify: $$x^2 + 2x + 1 = 0$$ Factor the equation: $$(x+1)^2 = 0$$ So, we have only one critical point: \(x = -1\).

Analyze the intervals around the critical point

We test the intervals around the critical point to determine the concavity of the function: Interval 1: \(x < -1\) \(x = -2 \Rightarrow f''(-2) = 24(-2)^2 + 48 (-2) + 24 = 96 - 96 = 0\) As f''(-2) = 0, it does not help us to determine concavity. Let's check x = -3. \(x = -3 \Rightarrow f''(-3) = 24(-3)^2 + 48 (-3) + 24 = 216 - 144 = 72 > 0\) So, the function is concave up on this interval. Interval 2: \(x > -1\) \(x = 0 \Rightarrow f''(0) = 24(0)^2 + 48(0) + 24 = 24 > 0\) The function is also concave up on this interval. Since the second derivative does not change its sign around the critical point, there are no inflection points. In conclusion, the function is concave up on the intervals \((-\infty, -1)\) and \((-1, \infty)\). There are no inflection points.

Key concepts

Concavity

To understand concavity, picture a curved line. If it bends upwards like a smile, it's concave up. If it bends downwards, like a frown, it's concave down. In mathematical terms, concavity tells us whether the slope of a curve is increasing or decreasing.
To determine concavity, we use the second derivative of a function:

• If the second derivative is greater than zero ( \( f''(x) > 0 \) ), the function is concave up in that interval.
• If the second derivative is less than zero ( \( f''(x) < 0 \) ), the function is concave down.

By analyzing the sign of the second derivative across intervals, you identify these concavity changes. Understanding this helps predict how a function behaves, making it useful in various applications like economics and physics.

Inflection Points

An inflection point is where a function changes concavity. Imagine driving on a hilly road: an inflection point is where you go from uphill to downhill or vice versa.
To find inflection points, examine where the second derivative is either zero or undefined, as these marks potential places for a change in concavity. However, simply having a zero isn't enough—it must actually swap signs on either side.
In the exercise, we found the second derivative to have a critical point at\( x = -1 \). But since there was no sign change from negative to positive or vice versa around this point, there are no inflection points for the given function. This highlights the importance of testing intervals and not just relying on where the derivative equals zero.

Second Derivative

The second derivative is a derivative of the derivative—a bit like measuring the change in acceleration in physics. It helps us understand how the slope of a curve is changing.
For the function\( f(x)=2x^{4}+8x^{3}+12x^{2}-x-2 \), the first derivative is the rate of change of the function, and the second derivative gives the rate at which this change is happening.
The steps for calculating the second derivative are simple:

• Find the first derivative by differentiating each term.
• Differentiate the first derivative to get the second derivative.

In our case, the second derivative\( f''(x) = 24x^2 + 48x + 24 \) was derived to determine behavior in various intervals. By evaluating this expression, you can tell where the function is concave up or down.

Quadratic Equations

A quadratic equation is a polynomial of the form \( ax^2 + bx + c = 0 \). It's essential when dealing with the second derivative, especially to find critical points.
Quadratics have real-life applications, like figuring out projectile paths or optimizing areas and volumes.
In the solution, the second derivative\( 24x^2 + 48x + 24 = 0 \) was simplified to\( x^2 + 2x + 1 = 0 \). This is a quadratic that can be factored to\((x+1)^2 = 0\), giving a solution\( x = -1 \). This technique of simplifying and solving quadratics helps find critical points and explore functions' behavior at those points. It's a valuable skill for breaking down more complex algebra and calculus problems into manageable parts.