Calculus Cheat Sheet
Extrema
Absolute Extrema Relative (local) Extrema
1. x = c is an absolute maximum of f (x) if 1. x = c is a relative (or local) maximum of f (x)
f (c) ≥ f (x) for all x in the domain. if f (c) ≥ f (x) for all x near c.
2. x = c is an absolute minimum of f (x) if 2. x = c is a relative (or local) minimum of f (x)
f (c) ≤ f (x) for all x in the domain. if f (c) ≤ f (x) for all x near c.
Fermat’s Theorem 1st Derivative Test
If f (x) has a relative (or local) extrema at x = c, If x = c is a critical point of f (x) then x = c is
then x = c is a critical point of f (x). 1. a relative maximum of f (x) if f 0 (x) > 0 to the
left of x = c and f 0 (x) < 0 to the right of x = c.
Extreme Value Theorem
If f (x) is continuous on the closed interval [a, b] then 2. a relative minimum of f (x) if f 0 (x) < 0 to the
there exist numbers c and d so that, left of x = c and f 0 (x) > 0 to the right of x = c.
1. a ≤ c, d ≤ b, 3. not a relative extrema of f (x) if f 0 (x is the
2. f (c) is the absolute maximum in [a, b], same sign on both sides of x = c.
3. f (d) is the absolute minimum in [a, b].
2nd Derivative Test
Finding Absolute Extrema If x = c is a critical point of f (x) such that f 0 (c) = 0
then x = c
To find the absolute extrema of the continuous
function f (x) on the interval [a, b] use the following 1. is a relative maximum of f (x) if f 00 (c) < 0.
process. 2. is a relative minimum of f (x) if f 00 (c) > 0.
1. Find all critical points of f (x) in [a, b].
3. may be a relative maximum, relative
2. Evaluate f (x) at all points found in Step 1. minimum, or neither if f 00 (c) = 0.
3. Evaluate f (a) and f (b).
Finding Relative Extrema and/or
4. Identify the absolute maximum (largest
Classify Critical Points
function value) and the absolute minimum
1. Find all critical points of f (x).
(smallest function value) from the
2. Use the 1st derivative test or the
evaluations in Steps 2 & 3.
2nd derivative test on each critical point.
Mean Value Theorem
If f (x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) then there is a
f (b) − f (a)
number a < c < b such that f 0 (c) = .
b−a
Newton’s Method
f (xn )
If xn is the nth guess for the root/solution of f (x) = 0 then (n + 1)st guess is xn+1 = xn − provided
f 0 (xn )
f 0 (xn ) exists.
© Paul Dawkins - https://tutorial.math.lamar.edu