Each card is a single, complete idea: what it is, the formulas you need, and one worked example. Built to be re-read.
Critical points occur where f′(x)=0 or is undefined. Use sign changes of f′ to classify them.
f′′>0⟹ concave up; f′′<0⟹ concave down. Inflection at sign changes of f′′.
If f is continuous on [a, b] and differentiable on (a, b), some interior point has instantaneous rate equal to the average rate.
Local (relative) extrema are any peaks or valleys; absolute (global) extrema are the single highest and lowest points on the whole interval. Concave up looks like a U (rate of change increasing), concave down like an upside-down U (rate of change decreasing). A point of inflection is where concavity actually changes.
If f is continuous on a closed interval [a, b], then f must attain both an absolute maximum and an absolute minimum somewhere on that interval. This is the guarantee that makes the Candidates Test valid.
All three are existence theorems: verify the hypotheses in writing, then state what must exist. IVT gives a y-value, MVT gives a slope, EVT gives extreme points.
At a critical point where f′(c)=0, the sign of f′′(c) classifies it by concavity. If f′′(c)=0 the test is inconclusive — fall back on the first derivative test.
On a closed interval, absolute extrema can only occur at critical points or endpoints. Evaluate the original function f (not f') at every candidate and compare the outputs.
Read behavior of f off the graph of f'. Height of f' gives the sign (increasing/decreasing), and the slope of f' gives concavity. Relative extrema of f sit where f' crosses the x-axis; inflection points of f sit where f' turns around.
Write a primary equation for the quantity to maximize or minimize, use the constraint to reduce it to one variable, then find and test critical points. Always answer the question that was asked and justify why your answer is the extreme value.