Using Higher Derivatives to Determine the Nature of Critical Points

Oct 13, 2025 by talynt.org 34 views

Introduction

In calculus, a critical point of a function is a point where the derivative of the function is zero or undefined. These points are crucial in understanding the behavior of the function, as they can indicate local maxima, minima, or points of inflection. In this article, we will find all critical points of the function $f(x) = x^4 - x^2$ and use higher derivatives to determine whether each of these points is a local maximum, minimum, or an inflection point.

Finding Critical Points

To find the critical points of the function $f(x) = x^4 - x^2$ , we need to find the values of $x$ where the derivative of the function is zero or undefined. The derivative of the function is given by:

f'(x) = 4x^3 - 2x

We can set the derivative equal to zero and solve for $x$ to find the critical points:

4x^3 - 2x = 0

2x(2x^2 - 1) = 0

x = 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}

Therefore, the critical points of the function $f(x) = x^4 - x^2$ are $x = 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$ .

Using Higher Derivatives to Determine the Nature of Critical Points

To determine whether each of these points is a local maximum, minimum, or an inflection point, we need to use higher derivatives. The second derivative of the function is given by:

f''(x) = 12x^2 - 2

We can evaluate the second derivative at each of the critical points to determine the nature of the point:

At $x = 0$ , $f''(0) = -2 < 0$ , so $x = 0$ is a local maximum.
At $x = \frac{1}{\sqrt{2}}$ , $f''(\frac{1}{\sqrt{2}}) = 12(\frac{1}{\sqrt{2}})^2 - 2 = 5 > 0$ , so $x = \frac{1}{\sqrt{2}}$ is a local minimum.
At $x = -\frac{1}{\sqrt{2}}$ , $f''(-\frac{1}{\sqrt{2}}) = 12(-\frac{1}{\sqrt{2}})^2 - 2 = 5 > 0$ , so $x = -\frac{1}{\sqrt{2}}$ is a local minimum.

Conclusion

Inflection Points

To find the inflection points of the function, we need to find the values of $x$ where the second derivative is zero or undefined. The second derivative is given by:

f''(x) = 12x^2 - 2

We can set the second derivative equal to zero and solve for $x$ to find the inflection points:

12x^2 - 2 = 0

12x^2 = 2

x^2 = \frac{1}{6}

x = \pm \frac{1}{\sqrt{6}}

Therefore, the inflection points of the function $f(x) = x^4 - x^2$ are $x = \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}$ .

Conclusion

In conclusion, the critical points of the function $f(x) = x^4 - x^2$ are $x = 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$ . Using higher derivatives, we have determined that $x = 0$ is a local maximum, while $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$ are local minima. The inflection points of the function are $x = \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}$ .

Graphical Representation

The graph of the function $f(x) = x^4 - x^2$ is a quartic function that has a local maximum at $x = 0$ and two local minima at $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$ . The graph also has two inflection points at $x = \frac{1}{\sqrt{6}}$ and $x = -\frac{1}{\sqrt{6}}$ .

Conclusion

Introduction

In our previous article, we found the critical points of the function $f(x) = x^4 - x^2$ and used higher derivatives to determine whether each of these points is a local maximum, minimum, or an inflection point. In this article, we will answer some frequently asked questions about the critical points of the function.

Q: What are the critical points of the function $f(x) = x^4 - x^2$ ?

A: The critical points of the function $f(x) = x^4 - x^2$ are $x = 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}$ .

Q: How do you determine the nature of the critical points?

A: To determine the nature of the critical points, we need to use higher derivatives. The second derivative of the function is given by:

f''(x) = 12x^2 - 2

We can evaluate the second derivative at each of the critical points to determine the nature of the point.

Q: What is the nature of the critical point $x = 0$ ?

A: At $x = 0$ , $f''(0) = -2 < 0$ , so $x = 0$ is a local maximum.

Q: What is the nature of the critical points $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$ ?

A: At $x = \frac{1}{\sqrt{2}}$ , $f''(\frac{1}{\sqrt{2}}) = 12(\frac{1}{\sqrt{2}})^2 - 2 = 5 > 0$ , so $x = \frac{1}{\sqrt{2}}$ is a local minimum. Similarly, at $x = -\frac{1}{\sqrt{2}}$ , $f''(-\frac{1}{\sqrt{2}}) = 12(-\frac{1}{\sqrt{2}})^2 - 2 = 5 > 0$ , so $x = -\frac{1}{\sqrt{2}}$ is a local minimum.

Q: What are the inflection points of the function $f(x) = x^4 - x^2$ ?

A: The inflection points of the function $f(x) = x^4 - x^2$ are $x = \frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}$ .

Q: How do you graph the function $f(x) = x^4 - x^2$ ?

A: The graph of the function $f(x) = x^4 - x^2$ is a quartic function that has a local maximum at $x = 0$ and two local minima at $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$ . The graph also has two inflection points at $x = \frac{1}{\sqrt{6}}$ and $x = -\frac{1}{\sqrt{6}}$ .

Q: What is the significance of the critical points and inflection points of the function $f(x) = x^4 - x^2$ ?

A: The critical points and inflection points of the function $f(x) = x^4 - x^2$ are important in understanding the behavior of the function. They can indicate local maxima, minima, or points of inflection, which can be useful in various applications such as optimization, physics, and engineering.