Critical Points Calculator

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Critical Points Lesson

Lesson Contents

What are Critical Points?

A critical point is a point on a given domain of a function where the function’s derivative is either zero or undefined, and the function itself exists at that point.

How to Find the Critical Points of a Function?

Critical points x = c are found under the following conditions:

1.)  f ‘(c) equals zero  OR  f ‘(c) is undefined

2.)  f(c) exists

Where c is the critical point that satisfies both conditions, f ‘(c) is the derivative of the input function f(x) evaluated at x = c, and f(c) is the input function f(x) evaluated at x = c.

Steps for finding the critical points of a given function f(x):

1.) Take derivative of f(x) to get f ‘(x)

2.) Find x values where f ‘(x) = 0 and/or where f ‘(x) is undefined

3.) Plug the values obtained from step 2 into f(x) to test whether or not the function exists for the values found in step 2

4.) The x values found in step 2 where f(x) does exist can be taken as critical points since the function exists at these points and they lie within the domain of our function f(x)

Example Problem 1

$$\begin{align}& \text{1.) Find Critical Points of} f(x) \\ \\ & \hspace{3ex} \text{Critical Points } \: x = c \: \text{ are found where:} \\ \\ & \hspace{3ex} f(c) \text{ exists and one of the following conditions are met:} \hspace{15ex} \\ \\ & \hspace{3ex} f'(c) = 0 \; \text{ or } \; f'(c) \text{ is undefined}\\ \\ & \text{2.) Given } f(x) = \frac{x^3}{3}-16x\text{, find } f'(x) \text{:} \\ \\ & \hspace{3ex} f'(x) ={x}^{2} – 16\\ \\ & \text{3.) Find } x \text{ values such that } f'(x) = 0 \text{ or } f'(x) \text{ is undefined } \\ \\ & \hspace{3ex} \text{to get provisional critical points } c \text{ :} \\ \\ & \hspace{3ex} \Rightarrow \hspace{2ex} {x}^{2} – 16 = 0 \\ \\ & \hspace{3ex} \Rightarrow \hspace{2ex} \text{After solving for } x \text{ we get } x =4,-4\\ \\ & \hspace{3ex} \Rightarrow \hspace{2ex} c =4,-4\\ \\ & \text{4.) Since our domain is} \hspace{1ex} – \infty < x < \infty \hspace{1ex} \text{(all real numbers), and our } c \text{ values} \\ \\ & \hspace{3ex} \text{are real numbers, we will plug all of these } c \text{ values into } f(x) \text{ to get } f(c) \text{.}\\ \\ & \text{5.) Plugging our real } c \text{ values into } f(x) \text{ to get } f(c) \text{, we get:}\\ \\ & \hspace{3ex} f(c_{1}) = \frac{(4)^3}{3}-16(4) = -42.666666666666664\\ \\ & \hspace{3ex} \text{Since } f(x) \text { exists at } c_{1}, \text{ there is a critical point at } c_{1} \text{ = } 4\\\\ \\ & \hspace{3ex} f(c_{2}) = \frac{(-4)^3}{3}-16(-4) = 42.666666666666664\\ \\ & \hspace{3ex} \text{Since } f(x) \text { exists at } c_{2}, \text{ there is a critical point at } c_{2} \text{ = } -4\\\end{align}$$

Example Problem 2

$$\begin{align}& \text{1.) Find Critical Points of} f(x) \\ \\ & \hspace{3ex} \text{Critical Points } \: x = c \: \text{ are found where:} \\ \\ & \hspace{3ex} f(c) \text{ exists and one of the following conditions are met:} \hspace{15ex} \\ \\ & \hspace{3ex} f'(c) = 0 \; \text{ or } \; f'(c) \text{ is undefined}\\ \\ & \text{2.) Given } f(x) = \frac{1}{3x}-\frac{1}{6}x^2\text{, find } f'(x) \text{:} \\ \\ & \hspace{3ex} f'(x) =- \frac{1}{3} \cdot x – \frac{1}{3 \cdot {x}^{2}}\\ \\ & \text{3.) Find } x \text{ values such that } f'(x) = 0 \text{ or } f'(x) \text{ is undefined } \\ \\ & \hspace{3ex} \text{to get provisional critical points } c \text{ :} \\ \\ & \hspace{3ex} \Rightarrow \hspace{2ex} – \frac{1}{3} \cdot x – \frac{1}{3 \cdot {x}^{2}} = 0 \\ \\ & \hspace{3ex} \Rightarrow \hspace{2ex} \text{After solving for } x \text{ we get } x =-1,\left(\frac{1}{2}\right) \cdot i \cdot \sqrt{3} + \frac{1}{2},\left(- \frac{1}{2}\right) \cdot i \cdot \sqrt{3} + \frac{1}{2},0\\ \\ & \hspace{3ex} \Rightarrow \hspace{2ex} c =-1,\left(\frac{1}{2}\right) \cdot i \cdot \sqrt{3} + \frac{1}{2},\left(- \frac{1}{2}\right) \cdot i \cdot \sqrt{3} + \frac{1}{2},0\\ \\ & \text{4.) Since our domain is} \hspace{1ex} – \infty < x < \infty \hspace{1ex} \text{(all real numbers), we will not plug} \\ \\ & \hspace{3ex} \text{the imaginary values back into } f(x) \text{ to find } f(c) \text{. Instead, we will proceed} \\ \\ & \hspace{3ex} \text{with only the } c \text{ values that are real numbers.}\\ \\ & \text{5.) Plugging our real } c \text{ values into } f(x) \text{ to get } f(c) \text{, we get:}\\ \\ & \hspace{3ex} f(c_{1}) = \frac{1}{3(-1)}-\frac{1}{6}(-1)^2 = -0.5\\ \\ & \hspace{3ex} \text{Since } f(x) \text { exists at } c_{1}, \text{ there is a critical point at } c_{1} \text{ = } -1\\\\ \\ & \hspace{3ex} f(c_{2}) = \frac{1}{3(0)}-\frac{1}{6}(0)^2 = \text{undefined} \\ & \hspace{3ex} \text{No critical point exists at this c value since } f(c) \text{ is not defined at this point.}\\\end{align}$$

How the Calculator Works

The Critical Points Calculator was developed using HTML, CSS, and JavaScript. The HTML portion of the code creates the framework of the calculator. This framework designates various important entities such as the calculator space itself, answer box, and the solution box.

CSS is then used to provide custom styling to the calculator and answer elements so that it is easy to read and use as needed. Everything from the clickable calculator button styling to the automated scroll bars in the solution field are styled using CSS.

For calculator functionality, JavaScript is used. This includes physical button presses, deriving the user’s input function, and outputting the custom solution steps tailored to the user’s inputs. When all of these components come together, we get a smooth and satisfying calculator experience that is responsive to the user’s specific needs.

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