AB Practice Test Review

∫1/x² dx

*do NOT take ln x², this ONLY works for just 1/x* ∫x⁻² = -x⁻¹ + C

if f(x) = ln(x+4+e^-3x) then f'(0) =

*remember d/dx of e^x= e^x (d/dx x)* ANSWER: -2/5

ln 1

0

two vertices of a triangle are (1,0) and (5,0) and a third vertex is on the graph of y= ln(2x) - ½x + 5 for ½≤x≤8. What is the max area of triangle

1) *create formula* A = ½bh A = ½(4) (ln2x - ½x + 5) 2) find d/dx A' = 2 (1/2x) x 2 -1 A' = 2/x - 1 3) find CVs (set =0) x=2 4) sign chart (to test if x=2 is max) yes 5) *plug x=2 into og equation* ANSWER: 2ln4 + 8

whenever you are given dy/dx and you need to find d²y/dx² at a given x and y

1) *implicit* differentiation (leave dy/dx before every y) 2) plug in dy/dx as the value given in problem 3) be sure to plug in *y* value into equation

lim as x→3 of: [tan(x-3)] / [3e(^x-3) - x]

1) L'Hospital (take d/dx) sec²(x-3) / 3e(^x-3) - x 2) change to (1/cos x 1/cos) and plut in 3 ANSWER: 1/2

∫x²cos(x³)dx

1) always try u sub 1st when see composite function ANSWER: 1/3 sin x³ + C

f(x) = (x-1) (x²+2)³ what is f'(x)?

1) chain rule inside multiplication rule (x²+2)³ + 6x(x²+2)²(x-1) 2) factor (x²+2)² [(x²+2)+ 6x(x-1)] (x²+2)² [(x²+2)+ 6x²-6x] (x²+2)² [7x²-6x+2]

does trapezoidal sum over or underestimate?: 1) concave up graph 2) concave down graph

1) concave up: overestimate 2) concave down: underestimate

how to use regular trapezoidal rule (when all have same interval)

1) define number of intervals 2) define ∆x (base width of each interval) 3) ∆x/2 (y₀ + 2y₁ + 2y₂ + 2y₃ + y₄) (find y by plugging in interval values into function inside ∫)

volume using base/cross sections

1) determine shape (given) and write corresponding equation 2) graph -determine limits -determine which function bigger 3) plug functions into equation 4) take integral using limits

∫²₁ (x²-x-5) / (x+2)

1) divide out first [x-3] + [1/(x+2)] 2) integrate -4 + ln4 + 5/2 ln3 ANSWER: -3/2 ln 4/3

when given a graph and asked to relate f f' and f", what do you do

1) draw fun chart 2) find the symbol that matches the interval your talking about 3) relate it to other f on chart

let f be the function given by f(x) = 2xe^x. The graph of f is concave down when x...

1) find f"(x) f"(x) = 4e^x + 2xe^x 2) find when negative *e is ALWAYS positive 4x+2x=0 x=-2 ANSWER: x<-2

g(x) = x²+bx where b is a constant. If the line tangent to the graph of g at x=-1 is parallel to the line that contains the points (0,-2) and (3,4), what is the value of b?

1) find slope (4+2) / (3-0) m=2 2) set g(x) equation = slope and plug in -1 g(-1) = 2(-1) + b 2 = 2(-1) + b ANSWER: b=4

if you are given the graph of f'(x) with x and y intercepts and a coordinate set of f(x), how can you find f(1)?

1) find slope of f'(x) using coordinates from intercepts; put in b as y intercept 2) take integral of f'(x) equation 3) find C using f(x) points given 4) plug in x=1 into new equation

f(3)=15, f(6)=3, f'(3)=-8, and f'(6)=-2. g(x)=f⁻¹(x) for all x. What is the value of g'(3)?

1) formula g'(x) = 1/ f'(g(x)) g'(3) = 1/ f'(g(3)) 2) inverse means switch x and y f(6)=3 so g(3)=6 ANSWER: -1/2

if g'(x)>0 and g"(x)>0 for all real numbers of x, such that g(4)=12 and g(5)=18, what could be a possible value of g(6)?

1) fun chart *f' and f" is + so f is increasing and concave up 3) that means f is going up hill so must be increasing at greater rate than from g(4)-g(5) *find slope from g(4)-g(5) = 6 g(6) > 18+6 ANSWER: g(6) > 24

f(x) = {|x| / x for x≠0} {0 for x=0} what is the value of ∫³-₅ f(x) dx

1) graph (memorize) 2) shade portions from -5 to 3 3) (+rectangle above x axis) (-rectangle below x axis) ANSWER: -2

what is the total area of the regions between the curves y=6x²-18x and y=-6x from x=1 to x=3?

1) graph and sketch intersection areas 2) set up two different integrals (different limits) ∫²₁g(x)-f(x) + ∫³₂ f(x) -g(x) ANSWER: 12

what is the solution to the differential equation dy/dx = x²/y with initial condition y(3) = -2

1) identify problem is in criss cross format 2) get like terms together ydy = x²dx 3) integral ∫y dy = ∫x² dx y²/2 = x³/3 + C 4) solve for C by plugging in initial condition C = -7 5) get equation in terms of y= ANSWER: y = √[(2x³/3) - 14]

the surface area of a tank of water is A=12πh - πh² sq m. h is the depth in meters of the water in tank. When h=3m the depth of the water is decreasing at rate of 1/2 m/min. At that instant, what is the rate at which the area of the water's surface is decreasing w respect to time?

1) implicit differentiation dA/dt = 12π dh/dt - 2πh dh/dt 2) plug in h=3 and dh/dt=-1/2 ANSWER: -3π

f(1) = 2 and dy/dx = y³+3. What is the value of d²y/dx² at x=1

1) implicit differentiation 2) plug dy/dx = y³+3 into d²y/dx² = 3y² (dy/dx) 3) plug in y as 2 ANSWER: 132

finding limit by dividing coefficient of numerator by denominator ONLY works if

1) lim →∞ 2) leading coeff have same power of x

what do you do when your given a piecewise function and you have to find if two equations have a limit, are continuous, and differentiable

1) limit exists if intersect at the specified point 2) continuous if only have one y value for x value 3) probably not differentiable bc intersecting lines make corners or cusps

if f(x) = sin⁻¹x, then f'(√3/2) =?

1) plug (√3/2) into equation for d/dx of sin⁻¹x 2) square denominator components separately 3) 1/√¼ = 1/(½) = 2 ANSWER: 2

f(x) = x²+2x. What is d/dx (f(ln x))

1) plug in ln x into all x's in f(x) 2) take d/dx ANSWER: (2lnx +2) / x

if you know the area under each curve of function from -3 to 3, how would you find ∫f(x) + 1 from -3 to 3?

1) shift function up one 2) take area of rectangle difference between two functions (bh = 6(1) = 6) 3) add to og area: og: 2-2-2 = -2 -2+6 = 4 ANSWER: 4

the line x+y=k (k is constant), is tangent to graph of y = x²+3x+1. What is value of k

1) slope y=k-x m=-1 2) x y' = m y' = 2x+3= m -1 = 2x+3 x = -2 3) y y=x²+3x+1 y = (-2)²+3(-2)+1 y= -1 4) k k = x+y k = -3

a curve has a slope 2x+3 at each point (x,y) on the curve. What is the equation of the curve if it passes through the point (1,2)?

1) slope of curve = d/dx, so antidifferentiate ∫2x+3 = x²+3x+C 2) solve for C by plugging in (1,2) C = -2 ANSWER: x²+3x-2

f(2) = 3 and f'(2) = -5. g(x) = xf(x). What is the slope of the line tangent to the graph of g at x=2

1) slope of tangent = d/dx g'(x) = d/dx (x · f(x)) = 1 (f(x)) + (f'(x)) x = 3 - 5x = 3 - 5(2) = -7

second fundamental theorem of calculus

1) take antiderivative of "what it is" 2) upper limit of antideriv - lower limit of antideriv

let f be the function defined by f(x)=4x³-5x+3. What is the equation of the line tangent to the graph of f at the point where x=-1?

1) take d/dx of f'(x) 12x²-5 2) plug in -1 to get m m=7 3) solve for y by plugging -1 into og equation y=4 4) y-y₁ = m (x-x₁) y=7x+11

write an equation for tangent line for: f(x) = 3+ ∫₀ⁿ |t-1| dt at n=2

1) tangent line slope = d/dx d/dx (3) = 0 d/dx (∫₀ⁿ |t-1|) = |t-1| m = |t-1| = |2-1| = 1 2) find y f(2) = 3+ ∫₀² |t-1| = 3+1 = 4 3) y-y₁ = m(x-x₁) y-4 = 1(x-2) y = x+2

1) when can you NOT use regular trapezoidal rule? 2) what do you do?

1) when have different intervals 2) 1/2[(x₂-x₁) (y₁+y₂)] + [(x₃-x₂) (y₂+y₃)]...

the function f is defined by f(x) = x³+4x+2. If g is the inverse function of f and g(2)=0, what is the value of g'(2)?

1) work w/ inverse equation g'(x) = 1/f'(g(x)) g'(2) = 1/f'(g(2)) g'(2) = 1/f'(0) 2) find f'(x) f'(x) = 3x²+4 3) plug in f'(0) f'(0) = 4 ANSWER: g'(2) = 1/4

what is the slope of the line tangent to curve y = arctan(4x) at point at which x=1/4?

1) y' = slope of tangent line y' = (1 / 1+x²) *(d/dx x)* ANSWER: 2

√2

1.4

the function f is continuous and differentiable for -2<x<1. If f(-2) = 5 and f(1) = 4, which of the following could be false?: a) there is a c where f(c) = 0 b) there is a c where f'(c) = 0 c) there is a c where f(c) = 3 d) there is a c where f'(c) = 3 e) there exists a c where f(c) ≥ f(x) for all x

ANSWER: b a and c are intermediate value theorem d is mean value e is extreme value theorem

ln 0

DNE

when plugging your upper and lower limits into your antidifferentiated integral and the lower limit is 0...

DO THE WORK OUT (may not be 0)

what is the value of ∫₁⁴ f'(x) dx

FTC (d/dx and integral cancel out) f(4)-f(1)

lim as x→e of: [(x²⁰-3x) - (e²⁰-3e)] / [x-e]

MVT f'(x) = [f(x) - f(c)] / [x-c] given right part; find left (f'(x)) ANSWER: 20e¹⁹ - 3

average *rate of change* formula (and use)

ONLY for d/dx of function given ie: want avg velocity if given position want avg accel if given velocity

mean value theorem

[f(upper) - f(lower)] / [upper-lower]

when do you use u sub and when do you integrate by parts

always try u sub 1st when see composite function (btw no integration by parts on test ... hint hint)

when given values of x and f'(x), how can you tell when f(x) is decreasing?

any x value where f'(x) is 0 or negative

whenever asked about a horizontal asymptote of a graph, what is the limit?

as x→∞...

the velocity of a particle is: v(t) = 2-t² What is average velocity from time t=1 to t=3?

average *value* formula ANSWER: -7/3

f(x) = {(x²-4)/(x-2) if x≠2} {1 if x=2} which of following is/are true? A) f has limit at x=2 B) f is continuous at x=2 C) f is differentiable at x=2

breaks off at 2 so can NOT be B or C ANSWER: A (alternatively graph by simplifying (x²-4)/(x-2) to (x+2) and see that there is a limit/open dot at x=2 but approaches same point from both sides at x=2)

first fundamental theorem of calculus

derivative of integral is just "what it is" (switch out t for x)

how do you find limits if the highest x term in numerator and denominator are same?

divide coefficients

when given a table of f(x), f'(x), g(x), and g'(x) values, what is the value of h'(1) if h(x) = f(g(x)) *just give formula to find

f'(g(1)) x g'(1)

the following are points on the graph of f"(x) (0,5) ; (1,0) ; (2,-7) T or F?: the graph has a POI at x=1

false; cannot just reach 0 to be POI, must change signs (although changes signs in points before and after, we cannot say for sure it changes signs at this point)

if the line tangent to grpah of f at point (1,7) passes through point (-2,-2) then f'(1) =

find slope 3

how to find graph of f' given graph of f

fun chart

horizontal tangent vs vertical tangent

horizontal: y= vertical: x=

intermediate value theorem

if a continuous graph spans from x=1-4, and f(1) = -5 and f(4) = 5 then there is a value of c between 1-4 that -5<f(c)<5

what is the slope of the line tangent to the curve 3y²-2x² = 6-2xy at the point (3,2)?

implicit differentiation ANSWER: 4/9

a particle moves along x axis so that at t>0, its acceleration is a(t) = ln(1+2^t). If the velocity of particle is 2 at t=1, then velocity at t=2 is?

initial + ∫ = final 2 + ∫₁² (1 + 2^t) 3.346 (initial does not have to be at t=0; in this case it's at t=1)

compare the estimates of the area of a concave down curve found by: -integral -left riemann sum -right riemann sum -midpoint riemann sum -trapezoid sum

integral: exact left: overestimate right: underestimate midpoint: underestimate but>right trapezoid: underestimate but>midpoint (more accurate)

when given y = (in a slope field problem)

just think of what graph of function looks like

when given y = (in slope field)

just think of what graph of function looks like

turn this integral into a limit: ∫⁵₂ x² dx

lim (as x→∞) ∑ f(2+(3k/n))² 1/n

division d/dx rule

low (d high) - *[*high (d low)*]* / low²

can a limit exist if there is more than one y value for an x value?

no

does a function have to be continuous at a point for the limit to exist at that point?

no

if f' is decreasing does that mean f is decreasing?

not necessarily matches up with concave down graph of f could be decreasing or increasing

if f' is increasing does that mean f is increasing?

not necessarily matches up with concave up graph of f could be increasing or decreasing

if two variables on opposite sides of equation are proportional, will they be in numerator or denominator?

numerator

when just using regular u/ du sub, what do you need to remember to do?

plug in limits into u equation to find NEW LIMITS

when given dy/dx =

plug in points to find graph

when given dy/dx = (in a slope field problem)

plug in points to find graph

whenever you see a square root!....

remember ± sign

what do you do when using washer method and one function starts below x axis and one starts above on the same interval but you only want area above?

separate into two intervals/integrals (*be careful to only include the f(x) that is + in 1st interval) ex image: π∫²₀ xdx + π∫⁴₂ (x-(x-2)²)dx

dy/dx = x+1 / y f(0) = -2 What is f(x)?

separate, integrate, solve for C, solve for y -√x²+2x+4 (*bc og y is -)

is the function y = e^3x - 5x +7 a solution to the equation: y" - 3y' - 15 = 0

take y' and y" of first function plug into second function see if = 0 ANSWER: yes

extreme value theorem

there is a x value on interval (c) where f(c) ≥ f(x) for all values of x on interval

whenever taking the antiderivative of something always remember...

to solve for C!

is a riemann sum or trapezoidal sum more accurate for a curved graph?

trapezoidal closer to actual value

average *value* formula (and use)

want avg of function given ie: want avg position if given position want avg velocity if given velocity want avg accel if given accel

how do you find extrema of f(x) on graph of f'(x)

where crosses x axis

if f' is positive does that mean f is increasing?

yes

if there is a single dot above a continuous graph, does the graph still have a limit at the x value where the dot is?

yes function still reaches same place from both sides at that x bc continuous (dot does NOT matter)

when asked to find ∫ based on geometry...

you are finding area between curve and X AXIS! *shade the area before determining answer* (may actually have to subtract shape to find area)

how to find graph given 1 / 4-2 ∫₂⁴ f(t) dt = 1

∫f(t) = 2 take area under curve: which is 2?

how can you find the number of POIs from a graph of f"(x)

# of times crosses x axis (changes signs)