A critical number (or critical value) is a number “c” that is in the domain of the function and
- Makes the derivative equal to zero: f′(c) = 0, or
- Results in an undefined derivative (i.e. it’s not differentiable at that point): f′(c) = undefined.
Critical numbers indicate where a change is taking place on a graph. For example:
- An increasing to decreasing point (e.g. a local maximum), or
- A decreasing to increasing point (e.g. a local minimum).
The number “c” has to be in the domain of the original function (the one you took the derivative of). If the number isn’t in the domain (for example, if there is a removable discontinuity at x = 0), then that number isn’t a critical number. It’s for this reason (there might be a miniscule hole in the graph), that you can’t rely on a graph to find critical numbers. In general, you have to find them with algebra.
Critical points are the points on the graph of the function. Once you’ve found a critical value, substitute the x-value into the function to get your y-value. The point (x, y) is your critical point.
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How to find critical numbers
Finding critical numbers is relatively easy if your algebra skills are strong; Unfortunately, if you have weak algebra skills you might have trouble finding critical numbers. Why? Because each function is different, and algebra skills will help you to spot undefined domain possibilities like division by zero. If your algebra isn’t up to par—now is the time to restudy the old rules.
Example question 2: Find the critical numbers for the following function:
Step 2: Figure out where the derivative equals zero. This is where a little algebra knowledge comes in handy, as each function is going to be different.
For this particular function, the derivative equals zero when -18x = 0 (making the numerator zero), so one critical number for x is 0 (because -18(0) = 0).
Another set of critical numbers can be found by setting the denominator equal to zero; When you do that, you’ll find out where the derivative is undefined:
- (x2 – 9) = 0
- (x – 3)(x + 3) = 0
- x = ±3
Step 3: Plug any critical numbers you found in Step 2 into your original function to check that they are in the domain of the original function. For this function, the critical numbers were 0, -3 and 3. Let’s plug in 0 first and see what happens:
f(x) = 02⁄02-9 = 0. Therefore, 0 is a critical number.
For +3 or -3, if you try to put these into the denominator of the original function, you’ll get division by zero, which is undefined.
f(x) = 32⁄32-9 = 9/0. Therefore, 3 is not a critical number.
That means these numbers are not in the domain of the original function and are not critical numbers.
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Difeng, C. MATH 111 Calculus I : 4.1 Maximum and Minimum Values. Retrieved August 18, 2021 from: http://www.math.emory.edu/~dcai7/day15.html
Stephanie Glen. "Critical Numbers or Values (Points): How to Find Them" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/calculus-problem-solving/find-critical-numbers-values/
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