AP Calc Task Cards 1-50

Removable Discontinuity

the function is continuous everywhere except for a “hole” at x =c

Intermediate value theorem

if f(x) is a continuous function and a < b and there is a value with such that n is between f(a) and f(b) then there is a number c such that a <c<b & f(c) = w

jump disconttinuity

the function has a left and right hand limit, but they don’t equal each other

end behavior model

the function g(x) is:

a) a right end behavior model for f if and only if lim x approaching infinity f(x)/ g(x) = 1

b) a left end behavior model for f if and only if lim x approaching - infinity f(x)/ g(x) = 1

average rate of change on an interval

f(b) - f(a)/ b- a

squeeze theorem

if g(x) ≤ f(x) ≤ h(x) for all x ≠ c in some interval about c, and lim as x approaches c g(x) = lim as x approaches c h(x) = L

then, lim as x approaches c f(x)=L

instantaneous rate of change

lim as h approaches 0 f(x+h) - f(x)/h

limits: a function f(x) has a limit as x approaches c if & only if the right hand and left hand limits at c exist & are equal

lim as x approaches c f(x) = L = lim as x approaches c + f(x) = L = lim as x approaches c - f(x) = L

infinite discontinuity

the function value increases or decreases indefinitely as x approaches c from the left and right

continuity test

1) f(x) is defined at c, that is f(c) exists

2) f(x) approaches the same value from either side of c, lim as x approaches c f(x) exists

3) the value that f(x) approaches from each side of c is f(c); lim as x approaches c f(x) = f(c)

sum & difference rule

d/dx (u ± v) = du/dx ± dv/dx

The derivative

the function whose value is f’(x) = lim h approaches 0 f(x+h) -f(x)/h

Power Rule for Positive Integer Powers of x

d/dx(x^n)=nx^n-1

implication of differentiability

1. local linearity

2. continuity

Constant Multiple Rule

d/dx(cu) = cdu/dx

one sided derivatives & differentiability

A function y = f(x) is differentiable on a closed interval [a,b] if it has a derivative at every interior point of the interval and if the limits:

lim h approaches 0+ f(a+h) - f(a)/ h

lim h approaches 0- f(b+h) - f(b)/h

exist at the endpoints

d/dx(sec(x)) =

sec(x)tan(x)

d/dx csc(x) =

-csc(x) cot(x)

general equation for displacement

s(t) = ½ gt² +v0t+s0

g = 32ft/sec²

g= 9.8m/sec²

d/dx (tan(x))=

sec²(x)

d/dx (cot(x))=

-csc²(x)

d/dx (sin(x)) =

cos(x)

speed

speed = |v(t)| = |ds/dt|

d/dx (cos(x)) =

-sin(x)

d/dx (a^x)=

a^x * ln(a)

d/dx (e^x)

e^x

derivative of a constant function

df/dx = d/dx (c ) = 0

slope of a curve at a point

slope of a curve y = f(x) at the point p(a, f(a)) is the number:

m = lim h approaches 0 f(a+h) - f(a)/h

how f’(a) might fail to exist

1. corner

2. cusp

3. vertical tangent

4. discontinuity

derivative of a function at the point x = a

f’(a) = lim x approaches a f(x)-f(a)/x-a

d/dx (sin^-1x)

1/ √1-x²

d/dx (cot^-1x)

-1/1+x²

d/dx (sec^-1x)

1/ x√x²-1

d/dx (cos^-1x)

-1/√1-x²

d/dx (tan^-1x)

1/1+x²

implict differentiation process

1. differentiate both sides of equation with respect to x

2. collect the terms with dy/dx on one side of equation

3. factor out dy/dx

4. solve for dy/dx

d/dx (csc^-1x)

-1/ x√x²-1

chain rule

if f if is differentiable at point u=g(x), and g is differentiable at x,

• then the composite function y=f(u) for u=g(x) is differentiable at x and dy/dx=dy/du * du/dx

• the the compostie function (fog)(x) = f(g(x)) is differentiable at x and (fog)’(x)=f’(g(x)) * g’(x)

product rule

d/dx (uv) = udv/dx + vdu/dx

quotient rule

d/dx(u/v) = vdu/dx - udv/dx/v²

concavity

the graph of a differentiable function y=f(x) is

a) concave up on an interval I if y’ is increasing on I

b) concave down on an interval I if y’ is decreasing on I

First derivative test for local extrema

for a continuous function f(x) at a critical point c:

1) if f’ changes sign from positive to negative at c, then f has a local maximum value at c

2) if f’ changes sign from negative to positive at c, then f has a local minimum value at c

3) if f’ doesn’t change sign at c, then f has no local extreme value at c

determining where function are increasing or decreasing

let f be continuous on [a,b] & differentiable on (a,b)

1. if f’>0 at each point of (a,b) then f increases on [a,b]

2. if f’<0 at each point of (a,b) then f decreases on [a,b]

mean value theorem

if y=f(x) is continuous at every point of the closed interval [a,b] & differentiable at every point of its interior (a,b), then there is at least one point c in (a,b) at which the instantaneous rate of change equals the mean rate of change,

f’(c)= f(b)-f(a)/b-a

stationary point

a point in the interior of the domain of a function f at which f’=0

critical point

a point in the interior of the domain of a function at which f’=0 or f’ does not exist

extreme value theorem

if f is continuous on a closed interval [a,b], then f has both a max value and min value on the interval

d/dx (log_aU)

1/uln(a) * du/dx

derivative of an inverse function

(f^-1)’(x) = 1/ f’(f^-1(x))

d/dx(ln(u))

1/u * du/dx