AP Calculus BC Exam Review 100% Correct Answers!!!

limit of one/both functions are DNE - ANS evaluate the limit from the left and the right hand of the approaching value values match = limit exists squeeze theorem - ANS f(x) h(x) g(x) limit of f(x) = limit of g(x) = # by squeeze theorem limit of h(x) = # removeable discontinuity - ANS limit does not equal the point jump discontinuity - ANS limit from the left does not equal the limit from the right infinite discontinuity - ANS limit approaches +/- infinity proving continuity - ANS at x = # limit from the left = limit from the right = f(#) finding vertical asymptotes of the original function - ANS set denominator equal to 0 IVT - ANS continuous f(x) ___ f(x) - y values so there exists a c, x c x such that f(x) = y-value by the IVT limit definition 1 - ANS limit definition 2 - ANS slope of a normal line - ANS negative reciprocal of slope of tangent line power rule - ANS derivative of ln x - ANS 1/x proving differentiability - ANS limit of f'(x) from the left = limit of f'(x) from the right derivative of tan x - ANS sec^2 x derivative of sec x - ANS sec x tan x derivative of cot x - ANS -csc^2 x derivative of csc x - ANS - csc x cot x taking derivative of y - ANS remember to add dy/dx finding horizontal tangents - ANS numerator of the slope = 0 finding vertical tangents - ANS denominator of the slope = 0 derivative of inverse functions - ANS set original function equal to the x-value given in regards to the inverse function plug solved x into the derivative of the original function reciprocate the solved y-value derivative of the inverse function = reciprocated y-value derivative of inverse sine - ANS remember to take the derivative of the x derivative of inverse tangent - ANS derivative of exponentials - ANS derivative of logarithms - ANS MVT for derivatives - ANS f(x) is continuous and differentiable there exists a value c on (interval) such that f'(c) = average rate of change by the MVT extreme value theorem - ANS f(x) is continuous there is at least on minimum and one maximum over the close interval finding critical points - ANS find derivative set derivative (numerator and denominator) equal to 0 find x value plug x value into the original function to check that they exists justification for intervals on which the function is increasing/decreasing - ANS f'(x) 0 or f'(x) 0 justification for relative minimum/maximum using first derivative - ANS f'(x) goes from - to + f'(x) goes from + to - maximum - ANS y-value cadidates test - ANS tells you the absolute minimum/maximum use when given a closed interval find critical points plug in critical point and end points into the original function point of inflection - ANS f"(x) changes sign f'(x) goes from increasing to decreasing and vice versa