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Intermediate Value Theorem
If f(1)=-4 and f(6)=9, then there must be a x-value between 1 and 6 where f crosses the x-axis.
average rate of change
Slope of secant line between two points, use to estimate instantanous rate of change at a point.
Instantenous Rate of Change
Slope of tangent line at a point, value of derivative at a point
Formal definition of derivative
limit as h approaches 0 of [f(a+h)-f(a)]/h
Alternate definition of derivative
limit as x approaches a of [f(x)-f(a)]/(x-a)
increasing
When f '(x) is positive, f(x) is
decreasing
When f '(x) is negative, f(x) is
Relative minimum
When f '(x) changes from negative to positive, f(x) has a
Relative maximum
When f '(x) changes fro positive to negative, f(x) has a
concave up
When f '(x) is increasing, f(x) is
concave down
When f '(x) is decreasing, f(x) is
point of inflection
When f '(x) changes from increasing to decreasing or decreasing to increasing, f(x) has a
corner, cusp, vertical tangent, discontinuity
When is a function not differentiable
Product Rule
uv' + vu'
(uv'-vu')/v²
Quotient Rule
Chain Rule
f '(g(x)) g'(x)
Product Rule
y = x cos(x), state rule used to find derivative
Quotient Rule
y = ln(x)/x², state rule used to find derivative
Chain Rule
y = cos²(3x)
velocity is positive
Particle is moving to the right/up
velocity is negative
Particle is moving to the left/down
Speed
absolute value of velocity
y' = cos(x)
y = sin(x), y' =
y' = -sin(x)
y = cos(x), y' =
y' = sec²(x)
y = tan(x), y' =
y' = -csc(x)cot(x)
y = csc(x), y' =
y' = sec(x)tan(x)
y = sec(x), y' =
y' = -csc²(x)
y = cot(x), y' =
y = sin⁻¹(x), y' =
y' = 1/√(1 - x²)
y = cos⁻¹(x), y' =
y' = -1/√(1 - x²)
y = tan⁻¹(x), y' =
y' = 1/(1 + x²)
y = cot⁻¹(x), y' =
y' = -1/(1 + x²)
y' = e^x
y = e^x, y' =
y' = a^x ln(a)
y = a^x, y' =
y' = 1/x
y = ln(x), y' =
y' = 1/(x lna)
y = log (base a) x, y' =
critical points and endpoints
To find absolute maximum on closed interval [a, b], you must consider...
Mean Value Theorem
if f(x) is continuous and differentiable, slope of tangent line equals slope of secant line at least once in the interval (a, b)
f(x) has a relative minimum
If f '(x) = 0 and f"(x) > 0,
f(x) has a relative maximum
If f '(x) = 0 and f"(x) < 0,
Linearization
use tangent line to approximate values of the function
rate
derivative
left riemann sum
use rectangles with left-endpoints to evaluate integral (estimate area)
right riemann sum
use rectangles with right-endpoints to evaluate integrals (estimate area)
Trapezoidal Rule
use trapezoids to evaluate integrals (estimate area)
[(h1 - h2)/2]*base
area of trapezoid
definite integral
has limits a & b, find antiderivative, F(b) - F(a)
indefinite integral
no limits, find antiderivative + C, use inital value to find C
area under a curve
∫ f(x) dx integrate over interval a to b
Positive
area above x-axis is
Negative
area below x-axis is
average value of f(x)
= 1/(b-a) ∫ f(x) dx on interval a to b
g'(x) = f(x)
If g(x) = ∫ f(t) dt on interval 2 to x, then g'(x) =
Fundamental Theorem of Calculus
∫ f(x) dx on interval a to b = F(b) - F(a)
To find particular solution to differential equation, dy/dx = x/y
separate variables, integrate + C, use initial condition to find C, solve for y
To draw a slope field,
plug (x,y) coordinates into differential equation, draw short segments representing slope at each point
zero
slope of horizontal line
Undefined
slope of vertical line
methods of integration
substitution, parts, partial fractions
use substitution to integrate when
a function and it's derivative are in the integrand
use integration by parts when
two different types of functions are multiplied
∫ u dv =
uv - ∫ v du
use partial fractions to integrate when
integrand is a rational function with a factorable denominator
dP/dt = kP(M - P)
logistic differential equation, M = carrying capacity
logistic growth equation
P = M / (1 + Ae^(-Mkt))
y₁ + Δy = y
t = a, R(t) = y₁ find final value when t = b
s₁+ Δs = s
given v(t) and initial position t = a, find final position when t = b
given v(t) find displacement
∫ v(t) over interval a to b
given v(t) find total distance travelled
∫ abs[v(t)] over interval a to b
area between two curves
∫ f(x) - g(x) over interval a to b, where f(x) is top function and g(x) is bottom function
volume of solid with base in the plane and given cross-section
∫ A(x) dx over interval a to b, where A(x) is the area of the given cross-section in terms of x
volume of solid of revolution - no washer
π ∫ r² dx over interval a to b, where r = distance from curve to axis of revolution
volume of solid of revolution - washer
π ∫ R² - r² dx over interval a to b, where R = distance from outside curve to axis of revolution, r = distance from inside curve to axis of revolution
length of curve
∫ √(1 + (dy/dx)²) dx over interval a to b