How to Find the Extrema of a Function

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Last Updated: February 25, 2026 References

Oftentimes in calculus, you will need to find the extrema (relative minimums and maximums) of a function. Sometimes you will be allowed to graph it and visually figure out where the graph peaks and dips, but other times you won't be able to. This article will teach you how to calculate the extrema using the first and second derivative tests.

Part 1

Part 1 of 2:

First Derivative Test

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1
Derive your function once. ^{[1]
X
Research source} Use one or a combination of derivative rules to derive your function.
- For example, you use the power rule to derive each term of the function $f(x) = x^3 - 1.5x^2 - 6x + 4$ . $f'(x) = 3x^2 - 3x - 6$ .
2
Find the x-values for which the derivative function equals zero or undefined. This also means you are finding the x values for which the slope of the function is zero or undefined. ^{[2]
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Research source} A function is undefined where there are asymptotes or when the denominator of a fraction equals 0, so you should include those as potential critical points as well. ^{[3]
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Research source}
- $3x^2 - 3x - 6$ is a polynomial we are able to factor. $f'(x) = 3(x^2 - x - 2)$ . The fully factored function is $f'(x) = 3(x+1)(x-2)$ .
- $3(x+1)(x-2) = 0$ ; the solutions are at x = -1 and x = 2. There are no x-values for which this quadratic function is undefined, meaning these are the critical points of this function.
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3
Create intervals between the critical points. Create one more interval than critical point you have. For example, if you have 3 critical points, you create 4 intervals. Place your intervals into a table.
- For this example, we have 2 critical points, so we create 3 intervals. The first one is between negative infinity and -1, the second one is between -1 and 2, and the last one is between 2 and positive infinity.
- This can be written in interval notation: (-∞, -1] U [-1, 2] U [2, ∞).
4
Find one number in each interval and substitute them into $f'(x)$ . This is to find out the values of $f'(x)$ $f'(x)$ in each interval. Put the $f'(x)$ $f'(x)$ values into your table.
- (-∞, -1]: x = -2. After substituting, $f'(x)$ is 12.
- [-1, 2]: x = 0. After substituting, $f'(x)$ is -6.
- [2, ∞): x = 3. After substituting, $f'(x)$ is 12.
5
Notice the sign changes of $f'(x)$ . A sign change for the derivative function indicates a relative maximum or minimum. You can determine this by looking at the left interval and right interval of a critical point.^{[4]
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Research source} If $f'(x)$ $f'(x)$ switches signs from the left interval to the right interval, there is a relative maximum or minimum.
- Relative maximum: $f'(x)$ changes from positive (+) to negative (-).
- Relative minimum: $f'(x)$ changes from negative (-) to positive (+).
- The point is not a relative maximum nor minimum if $f'(x)$ does not change signs ( $f'(x)$ is positive or negative over both intervals a critical point is located in).
- For this example, $f'(x)$ changed from positive (+12) to negative (-6) around x = -1. This means x = -1 is a relative maximum. $f'(x)$ also changed from negative (-6) to positive (+12) around x = 2. This means x = 2 is a relative minimum.
6
Verify your extrema with a graph. You can hand-draw the function, put the equation into a handheld graphing calculator, or use an online tool like Desmos to verify you got the correct relative maximums and minimums.

Part 2

Part 2 of 2:

Second Derivative Test

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1
Find your critical points. You will need to find your critical points like you did for the First Derivative Test. Derive your function once and find the x-values for which it equals 0 or undefined.
2
Derive your function twice. ^{[5]
X
Research source} Use one or a combination of derivative rules to derive your function. Do this one more time to get $f''(x)$ $f''(x)$ .
- For example, you use the power rule to derive each term of the function $f(x) = x^3 - 1.5x^2 - 6x + 4$ . $f'(x) = 3x^2 - 3x - 6$ .
- Use the power rule to derive $f'(x) = 3x^2 - 3x - 6$ again; it becomes $f''(x) = 6x - 3$ .
3
Substitute your critical points into the 2nd derivative. ^{[6]
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Research source} Plug in the critical points you found into $f''(x)$ $f''(x)$ to check if there are any sign changes.
- The critical points for $f(x) = x^3 - 1.5x^2 - 6x + 4$ are x = -1 and x = 2.
- After plugging in -1, $f''(x) = -9$ .
- After plugging in 2, $f''(x) = 9$ .
4
Notice the signs of $f''(x)$ . According to the Second Derivative Test, when $f''(x)$ $f''(x)$ is a positive number at a critical point, the graph goes concave up and the point is a relative minimum. When $f''(x)$ $f''(x)$ is a negative number, the graph goes concave down and the point is a relative maximum. ^{[7]
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Research source}
- This can also be stated as, "If $f''(x) > 0$ , then f has a relative minimum at x = c." and "If $f''(x) < 0$ , then f has a relative maximum at x = c."
- For this example, x = -1 is a relative maximum because $f''(x)$ is negative (-9) at that value. x = 2 is a relative minimum because $f''(x)$ is positive (+9) at that value.
5
Verify your extrema with a graph. You can hand-draw the function, put the equation into a handheld graphing calculator, or use an online tool like Desmos to verify you got the correct relative maximums and minimums.