Calculus AB U3

EVT, MVT, Rolle's, 1st derivative, 2nd derivative, L'hopitals

extreme value theorem (EVT)

• f is continuous on an interval containing f(a) → contains an absolute min/max value f(d)

how to do?

1. find f’(x) and critical numbers of f

• → critical numbers are when f’(x)=0 or f’(x)=DNE but remember that f(x) has to be cont. on that interval

2. evaluate f at every critical number (plug in values)

3. biggest number is max and smallest is min

mean value theorem (MVT)

• f(x) is continuous on [a,b] and differentiable —> at least 1 point f’( c ) = AROC or slope of tangent at c = AROC

  • remember that AROC is change in y over change in x

  • for real life applications this can be seen as distance (y) and time (x) which is velocity

Rolle’s theorem

• f(x) continuous and differentiable on [a,b]

• f(a)=f(b)

THEN…

• at least one number ( c ), f’( c ) =0

• f(x) has a HA at ( c )

• c is a critical number

increasing and decreasing

f’(x) > 0 POSITIVE → f(x) is increasing

f’(x) < 0 NEGATIVE → f(x) is decreasing

f’(x) = 0 CONSTANT → zero slope

first derivative test

• f is continuous and differentiable on an interval containing c ( where c is a critical point)

1. find f’(x)

2. find critical numbers

3. make a sign chart for f’(x) and see where its positive and neg

• f’(x) is (-) to (+) = local min

• f’(x) is (+) to (-) = local max

concavity

• CCU = f’ increasing + f”(x) positive

• CCD = f’ decreasing + f”(x) negative

inflection points

• F”(x)=0 or f”(x)=DNE

• changes signs (+) to (-) or (-) to (+) or from inc. to dec. vise versa

second derivative test

• F(x) is a function where f” exists on an interval containing c

1. if F’( c )=0 and F”( C ) <0 NEGATIVE → RELATIVE MAX

2. if F’( c )=0 and F”( C ) >0 POSITIVE → RELATIVE MIN