AB 1: Find lim x a f ( x)
AB 1 Answer: Step 1: Find f ( a). If you get a zero in the denominator, Step 2: Factor numerator and denominator of f ( x). Do any cancellations and go back to Step 1. If you still get a zero in the denominator, the answer is either, -, or does not exist. Check the signs of lim f ( x ) and lim f ( x ) for equality. x a x a +
Find lim x AB 2: f ( x) and lim x f ( x)
Denominator: lim x Numerator: lim x AB 2 Answer: Express f ( x) as a fraction. Determine location of the highest power: f ( x) = lim f ( x) = 0 x ( ) = ± (plug in large number) f x Both Numerator and Denominator: ratio of the highest power coefficients
AB 3: Determine if f ( x) is continuous at x = a
AB 3 Answer: 1) lim x a f ( x) exists: lim x a 2) f ( a) exists 3) lim f ( x) = f ( a) x a f ( x ) = lim x a + f ( x )
AB 4: Find the average rate of change of f on [a, b]
AB 4 Answer: Avg. rate of change = ( ) f ( a) f b b a
AB 5: Find the instantaneous rate of change of f at x = a
AB 5 Answer: Instantaneous rate of change = f ( a)
AB 6: The meaning and formal definition of the derivative of f ( x)
AB 6 Answer: Derivative: A formula for the slope of a tangent line to f f ( x) = lim h 0 ( ) f ( x) f x + h h
AB 7: Power rule for Derivatives
AB 7 Answer: d dx xn = nx n 1 and d dx ( ) f x n ( ) = n f x n 1 f ( x)
AB 8: Product rule for derivatives
d dx f ( x )g x ( ) AB 8 Answer: = f x ( ) g ( x) + g( x) f ( x)
AB 9: Quotient rule for derivatives
d dx ( ) ( ) f x g x AB 9 Answer: = g x ( ) f ( x) f ( x) g ( x) g( x) 2
AB 10 : Find the derivative of f ( g( x) )
AB 10 Answer: ( ) = f The chain rule: d dx f g ( x ) ( g( x) ) g ( x)
AB 11: Find the equation of the tangent line to f at ( x 1, y ) 1
AB 11 Answer: Find slope m = f ( x 1 ). Then use point-slope equation: y y 1 = m( x x ) 1
AB 12: The line y = mx + b is tangent to the graph of f ( x) at ( x 1, y ) 1
AB 12 Answer: Two relationships are true: 1) the function f and the line share the same y-value at x 1 2) the function f and the line share the same slope at x 1 : m = f ( x ) 1
AB 13: d dx lnu [ ]
AB 13 Answer: d [ dx lnu ] = 1 u du dx
AB 14: d dx eu and d dx au
AB 14 Answer: d du dx eu = e u dx d dx au = a u ln a du dx
d dx sinu [ ], AB 15: d dx cosu [ ], d dx tanu [ ]
AB 15 Answer: d dx sinu d dx cosu [ ] = cosu du dx [ ] = sinu du dx d [ dx tanu ] = sec 2 u du dx
d dx sin 1 u, AB 16: d dx cos 1 u, d dx tan 1 u
AB 16 Answer: d dx sin 1 u = d dx cos 1 u = d dx tan 1 u = 1 1 u 2 1 1 u 2 1 u 2 +1 du dx du dx du dx
AB 17: Find the derivative to the inverse of f ( x) at x =a
AB 17 Answer: Follow this procedure: 1) Interchange x and y 2) Plug the x-value into this equation and solve for y 3) Use the equation in 1) to find dy dx implicitly 4) Plug the y-value you found in 2) to find dy dx
AB 18 : Find x-values of horizontal and vertical tangents to f
AB 18 Answer: ( ) as a fraction in reduced form ( x) = 0 ( x) = 0 Find and write f x Horizontal: Set numerator of f Vertical: Set denominator of f
AB 19: Determine if a piecewise function f x ( ) is differentiable at x = a where the function rule splits
AB 19 Answer: First be sure that f x Then take the derivative f x lim f x x a ( ) is continuous at x = a. ( ) of each piece and show that ( ) = lim f ( x) x a +
AB 20: The volume of a solid is changing at the rate of
AB 20 Answer: dv dt =...
AB 21: How fast the area of a triangle with legs a and b is changing
AB 21 Answer: A = 1 2 ab da dt = 1 2 a db dt + b da dt
AB 22: Given position function x t ( ), find the velocity and acceleration functions
v t AB 22 Answer: ( ) = x ( t) ( ) = v ( t) = x ( t) a t
AB 23: Given a particle s position function x( t), find the average velocity on [ t 1,t ] 2
AB 23 Answer: ( ) x( t ) 1 Avg. velocity = x t 2 t 2 t 1
AB 24: Given a particle s position function x t ( ), Find the instantaneous velocity at t = k
AB 24 Answer: Inst. velocity = v k ( ) = x ( k)
AB 25: Given a particle s position function x t ( ), find values of t when the particle is stopped
AB 25 Answer: ( ) = x ( t) = 0 v t
AB 26: Given a particle s position function x t ( ), find values of t when the particle is speeding up or slowing down
AB 26 Answer: Find v( t) and a( t). Set each equal to zero to find intervals where v( t) and a( t) are positive and negative. Particle speeding up when v and a have the same signs. Particle slowing down up when v and a have opposite signs.
AB 27: Show the intermediate value theorem for f holds on [a, b]
If f x AB 27 Answer: ( ) is continuous on (a, b) and k is a number between f ( a) and f ( b), then there is some number c between a and b such that f ( c) = k.
AB 28: Show that the mean value theorem holds for f ( x) on ( a, b)
AB 28 Answer: Show that f is continuous and differentiable on [a, b]. Then there is some value c on [a, b] such that f ( c) = f ( b ) f ( a) b a
AB 29: Find critical values of f ( x)
AB 29 Answer: ( ) as a fraction. Set numerator and Find and express f x denominator equal to zero and solve.
AB 30: Find interval(s) where f ( x) is increasing/decreasing
AB 30 Answer: Find critical values of f ( x). Make a sign chart to find sign of f ( x) in the intervals bounded by critical values. Positive means f ( x) is increasing, negative means f ( x)is decreasing.
AB 31: Find points of relative extrema of f ( x)
AB 31 Answer: Make a sign chart of f ( x). At x = c where the derivative switches from negative to positive, there is a relative minimum. When the derivative switches from positive to negative, there is a relative maximum. To actually find the point, find f ( c).
AB 32: Given a graph of f ( x), find intervals where f x ( ) is increasing/decreasing.
Make a sign chart of f f x AB 32 Answer: ( x) and determine the intervals where ( ) is positive (increasing) and negative (decreasing).
AB 33: Use the 2 nd derivative test
If f If f AB 33 Answer: If f ( c) = 0, then ( c) > 0, there is a relative minimum at x = c. c If f c ( ) < 0, there is a relative maximum at x = c. ( ) = 0, the test is inconclusive.
AB 34: Find inflection points of f ( x)
AB 34 Answer: ( ) as a fraction. Set both numerator and Find and express f x denominator equal to zero and solve. Make a sign chart for f ( x). Inflection points for f occur when f x ( ) changes sign.
AB 35: ( ) ( ) Determine whether the linear approximation for f x 1 + a overestimates or underestimates f x 1 + a
AB 35 Answer: Find slope m = f ( x i ). Then use point slope formula y y 1 = m( x x 1 ). Evaluate this line for y at x = x 1 + a. ( ) > 0, f is concave up at x 1 and the linear If f x 1 approximation is an underestimation for f ( x 1 + a). ( ) < 0, f is concave down at x 1 and the linear If f x 1 approximation is an overestimation for f ( x 1 + a).
AB 36: Find the absolute maximum and minimum value of f x ( ) on [ a, b]
AB 36 Answer: Use relative extrema techniques to find locations of relative max/mins. Evaluate f at these values. Also examine f ( a) and f ( b). The largest of these is the absolute maximum and the smallest of these is the absolute minimum.
AB 37: Find the minimum slope of f ( x) on [ a, b]
AB 37 Answer: Find the derivative of f ( x) = f ( x). Find critical values of f x Values of x where f x are potential locations for the minimum slope. Evaluate f ( x) at those values and also f a choose the least of these value. ( ) and make a sign chart of f ( ) switches from negative to positive ( ) and f ( b) and ( x).
AB 38: Power rule for integrals
AB 38 Answer: ( ) x n dx = xn+1 n +1 + C or f u n du = f ( u ) n+1 n +1 + C
AB 39: 1 u du
AB 39 Answer: 1 u du = ln u + C
AB 40: eu du
AB 40 Answer: e u du = e u + C
AB 41: sinudu, cosudu, sec 2 udu
AB 41 Answer: sinudu = cosu + C cosudu = sinu + C sec 2 udu = tanu + C
If f x AB 42: ( ) > 0, find the area under the curve f ( x) on the interval [a, b]
AB 42 Answer: A = b ( ) f x dx a
x AB 43: ( ) f t dt a
AB 43 Answer: The accumulation function accumulated area under f starting at some constant a and ending at some variable x.
AB 44: ( ) Equivalent expression for f x dx where a < b a b
a AB 44 Answer: ( ) f x dx = f x dx b b a ( )
AB 45: The Fundamental theorem of calculus
b ( ) f x dx = F b a AB 45 Answer: ( ) F( a) where F is an anti-derivative of f
AB 46: Given the value of F( a), find the value of F b where the anti-derivative of f is F ( )
AB 46 Answer: F( b) = F( a) + f ( x) b dx a
AB 47: Approximate b f ( x)dx with left and right Riemann sums a
Left: Right: b a n b a n AB 47 Answer: f x 0 ( ) + f ( x ) 1 +... + f ( x ) n 1 f x 1 ( ) + f ( x ) 2 +... + f ( x ) n
e AB 48: Given the table of values below with b midway between a and c and d midway between c and e, approximate f ( x)dx with a midpoint Riemann sum with two intervals. a a b c d e f a ( ) f ( b) f ( c) f ( d) f ( e)
AB 48 Answer: A ( c a) f ( b) + ( e c) f ( d)
AB 49: Approximate b f ( x)dx using a trapezoidal summation a
AB 49 Answer: A = b a 2n f ( x ) 0 + 2 f ( x ) 1 + 2 f ( x ) 2 +... + 2 f ( x ) n 1 + f x n This formula only works when the base of each trapezoid is the same. If not, calculate the areas of individual trapezoids. ( )
d dx AB 50: x ( ) f t dt a
AB 50 Answer: d dx x ( ) f t dt = f x a ( )
AB 51: d dx g( x) ( ) f t dt a
AB 51 Answer: d dx g( x) f ( t) dt = f ( g( x) ) g ( x). a
AB 52: The area between f ( x) and g( x)
AB 52 Answer: Find the intersections a and b of f x If f ( x) g x If g( x) f x b ( ) and g x ( ), A = f ( x) g( x) dx a b ( ), A = g( x) f ( x) dx a ( ).
AB 53: The line x = c that divides the area under f x ( ) on [a, b] into two equal areas
c a b AB 53 Answer: f ( x) dx = ( ) dx = b c f ( x) dx f x 2 f x a c a ( ) dx or
AB 54: The average value of f ( x) on [a, b]
AB 54 Answer: F avg = b ( ) f x dx a b a
AB 55: The average rate of change of F [ ] ( t) on t 1,t 2
AB 55 Answer: d dt t 2 t 1 F ( x) dx = F ( t ) 2 F t 1 t 2 t 1 t 2 t 1 ( )
AB 56: b a ( ) The meaning of R t dt
b a AB 56 Answer: This is the accumulated change of R t R ( t) dt = R( b) R( a) or R( b) = R a ( ) on [a, b] b a ( ) + R ( t) dt
AB 57: Given a water tank with g gallons at t 1, filled at the rate of F( t) gallons/min and emptied at the rate of E t [ ], ( ) gallons/min on t 1,t 2 a) find the amount of water in the tank at t = m minutes b) find the rate the water amount is changing at t = m minutes
AB 57 Answer: a) g + F t b) d dt m 0 m 0 ( ) E( t) F( t) E( t) dt dt = F( m) E( m)
AB 58: ( ) and initial position x 0 ( ) Given velocity function v t find the expression for x t ( ),
( ) = v t x t AB 58 Answer: ( )dt + C. Plug in x 0 ( ) to find C.
AB 59: Given velocity function v( t), find the average velocity on t 1,t 2 [ ]
AB 59 Answer: Avg. velocity = t 2 t 1 v( t) dt t 2 t 1
AB 60: Given velocity function v( t), find the change of position of a particle between t 1 and t 2.
AB 60 Answer: Displacement = t 2 v( t) dt t 1
AB 61: Given velocity function v( t), find the distance a particle travels between t 1 and t 2.
AB 61 Answer: Distance = t 2 t 1 v( t) dt
AB 62: Given velocity function v t ( ) and initial position x 0 find the greatest distance of the particle from the starting position on [ 0,t ] 1 ( ),
AB 62 Answer: Generate a sign chart of v( t) to find turning points. x( t) = v( t)dt + C. Plug in x( 0) to find C. Evaluate x( t) at all turning points on 0,t 1 one gives the maximum distance from x( 0). [ ] and find which
AB 63: Find the volume when the area under f x ( ) is rotated about the x-axis on the interval [a, b]
Radius = f x AB 63 Answer: ( ) :V = π f ( x) b 2 dx disk method a ( )
AB 64: Find the volume when the area between f x is rotated about the x-axis ( ) and g( x)
AB 64 Answer: Outside radius = f ( x). Inside radius = g( x). Establish the values of a and b, where f ( x) = g( x). b ( ) dx V = π f ( x) 2 g( x) 2 - (washer method) a
AB 65: ( ) and g( x) on [a, b], Given a base bounded by f x find the volume when the cross sections of the solid perpendicular to the x-axis are squares
AB 65 Answer: Base = f ( x) g( x) Area = base 2 = V = b a ( ) g( x) 2 f x ( ) g( x) 2 f x dx
AB 66: y is increasing proportionally to y
AB 66 Answer: dy dt = ky which translates to y = Cekt