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AP Calculus

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Section 1

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Continuity on closed Interval

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Calculus Math

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Mar 1, 2020

Cards (25)

Section 1

(25 cards)

Continuity on closed Interval

Front

f cont. on closed interval [a.b] if cont. on open interval (a,b) lim f(x)=f(a) and lim f(x)=f(b) x->a+ x->b-

Back

Quotient Rule

Front

d/dx (f(x)/ g(x)) = [(h(x) g'(x) - g(x) h'(x))]/ h(x)^2 LDH_HDL L^2

Back

Horizontal Asymptote

Front

line y=L is a horizontal asymptote of f if lim f(x)=L or lim f(x)=L x->oo x-> -oo

Back

Limits of Trig. Functions

Front

Back

Chain Rule

Front

d/dx [f(g(x))] = f'(g(x)) g'(x)

Back

Continuity/Differentiability of Inverse Functions

Front

f domain is an interval I f has inverse function 1)f const. on its domain-> f^-1 is const on domain 2)f inc ->f^-1 is inc 3)f dec->f^-1 is dec 4)f diff' bles on interval containing c and f'(c) =/= 0 -->f' diff' bles at f(c)

Back

Def. of Derivative

Front

f'(x) = lim f(x+Δx)-f(x) Δx->0 Δx

Back

Limit involving e

Front

lim (1+(1/x))^x =lim (x+1/x)^x=e x->oo x->oo

Back

Derivatives of Trig Function

Front

d/dx sinx=cosx d/dx cscx=- cscxcotx d/dx cosx=-sinx d/dx secx= secxtanx d/dx tanx=sec^2(x) d/dx cotx=-csc^2(x)

Back

Defn's of Inverse Trig Functions

Front

y=arc sin x iff sin y=x D: -1<_x<_1 R:-π/2<_y<_π/2 y=arc cos x iff cos y=x D: -1<_x<_1 R:0<_y<_π y=arc tan x iff tan y=x D: -oo<_x<_oo R:-π/2<_y<_π/2 y=arc cot x iff cot y=x D: -oo<_x<_oo R:0<_y<_π y=arc sec x iff sec y=x D: |x| _>1 R:0<_y<_π, y=/= π/2 y=arc csc x iff csc y=x D: |x| _>1 R:-π/2<_y<_π/2, y=/=0

Back

Def. of Derivative

Front

f'(c) = lim f(x)-f(c) x->c x-c

Back

Derivatives of Inverse Trig Functions

Front

Back

derivative of inverse function

Front

f diff'ble on I. f has inverse f'ng -->g diff'ble at any c for which f'(g(x))=/=0 (f^-1)'(x)= 1/f'(f^-1(x)) = 1/(dx/dy)

Back

Existence of Limit

Front

Let f be function and let c and L be real lim f(x) =L x-->c iff lim f(x)=L and lim f(x) =L x-->c+ x-->c-

Back

Squeeze Theorem

Front

If h(x)<_ f(x)<_ g(x) for all x in an open interval containing c, except possibly at c itself and if lim h(x) = L = lim g(x) x->c x->c then lim f(x) exists and is equal to L. x->c

Back

Derivatives of ln

Front

d/dx ln x= 1/x d/dx ln u = 1/u • u'

Back

Intermediate Value Theorem (IVT)

Front

If f is continuous on the closed interval [a,b] and k is any number between f(a) and f(b), then there is at least one number c in [a,b] such that f(c)=k N=k

Back

Def. of Tangent Line with Slope m

Front

f defined on an open interval containing c, and if the limit lim Δy / Δx = lim f(c + Δx) - f(c) = m Δx->0 Δx->0 Δx exists, then line passing through (c,f(c)) with slope m is tangent line to graph of f at (c,f(c))

Back

Differentiability Implies Continuity

Front

If f is differentiable at x=c, then f is continuous at x=c cont. ->Diff?

Back

L' Hopitals Rule

Front

Let f and g be functions that are differntialbel on openinterval (a,b) containing c, except possibly at c itself. Assume g'(x) =/= 0 for all x in (a,b) except at a c itself. If limit gets oo/oo or 0/0 then lim f(x)/g(x) = lim f'(x)/g'(x) x->c x->c provides limit on right exists

Back

Derivatives of e^x

Front

d e^x=e^x dx

Back

Continuity

Front

f continous at c if 3 conditions met _ 1) f(c) is defined | 2)lim f(x) exists | x->c > at a point 3) lim f(x)=f(c) | x->c | -

Back

Derivatives of Bases other than e

Front

d/dx a^x=(lna)a^x d/dx a^u=(lna)a^u • u' d/dx log(a) x = 1/(lna)x d/dx log(a) u = 1 •u'/ ((lna)u)

Back

Position/Velocity

Front

Avg. Velocity = Δs = change in distance Δt change in time Position function s(t) = 1/2 g(t^2 ) + vo + so speed = |v| so= initial height vo=initial velocity g=acceleration due to gravity -32ft/(sec^2) or -9.8m/s^2 Velocity function v(t)= lim s(t + Δt) - s(t) = s'(t) Δt->0 Δt

Back

Product Rule

Front

d/dx[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

Back