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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