Chart is key to doing this type of problem
f'(x)=4x^3-48x=0
4x(x^2-12)=0
x=0 or x^2-12=0
=>x^2=12
x=2+-(v--3)
f''(x)=12x^2-48=0
12(x^2-4)=0
12(x+2)(x-2)=0
x=+-2
-4 -3 -1 1 3 4
(-@@,-2V--3)|(-2V--3,-2)|(-2,0)|(0,2)|(2,2V--3)|(2V--3,@@)
f'(x)+- |(-) (+) (+) (-) (-) (+)
f''(x)+- | (+) (+) (-) (-) (+) (+)
inc/dec | dec inc inc inc dec inc
concave up/down|up up down down up up
sketch | U intersect(grk) U
noticing mirror like combinations of positive to negative values, whengl
f'(x)=4x^3-48x
=4x(x^2-12)
f'(-4)=4(-4)((-4)^2-12) will be negative
f'(-3)=4(-3)((-3)^2-12)positive
f'(-1)=4(-1)((-1)^2-12)positive
extrapolated application of number to variables
rel max @ (0,f(0))
rel. min.@ (2V--3, f(2V--3)) and -2V--3, f(-2V--3)) (2root3, -66) and (-2root3, -66)
f(x)=x^4-24x^2+80
f(0)=80
f(2V--3)=16*9-24(4*3)+80
146-288+80=0
=-146+80= -66
^^same as f(-2V--3)
Exam Prep
Relative Extreme
very similar to what we get on the test, parts a thru e
just fill in all the blanks
not too hard, just to remember what is going on
any specific questions or walkthrough?
some derivatives?
probably one that involves more than one rule
List out all the rules-2columns
1) first side:
Derivatives we know
f(x)=x^n
=polynomial
e^x or a^x
ln^x
sinx cosx
2) rules to use we can "break up functions"
product rule u(x)-v(x)
quotient rule: p(x)/q(x)
chain rule: u(v(x))
ex) f(x) = sin^2(x+1/x)
=(sin(x+1)/(x))^2
CHAIN RULE
u(x)=x^2
u'(x)=2x
v(x)=sin((x+1)/x)
chain rule again
g(x)=sinx
g'(x)=cosx
h(x)=(x+1)/x
h(x)=(x/x)+1/x)
h'(x)=x^-2
v'(x)=g'(h(x))*h'(x)
=cos((x+1)/x)(-x^-2)
=-cos((x+1)/x)/x^2
f'(x)=u'(v(x))*v'(x)
2(sin((x+1)/x)(-cos((x+1)/x)^2)
ex) f(x)=(x^2-2)/e^x^2
Quotient rule:
p(x)=x^2-2
p'(x)=2x
q(x)=e^x^2
CHAIN RULE: q=u(v(x))
u(x)=e^x
u'(x)=e^x)
v(x)=x^2
v'(x)=2x
q'(x)=e^x*(2x)
f'=(e^x^2*2x-(x^2-2)(2xe^x^2))/(e^x^2)^2
=(2xe^x^2)-(2x^3e^x^2)+(4xe^x^2)/(e^2x^2)
=(xe^x^2)(6-2x^2)/(e^x^2)^2
=x(6-2x^2)/(e^x^2)
f(x)=3xe^x+2
find where ifs inc/dec,
concave up/down
PRODUCT RULE
u(x)=3x
u'(x)=3
v(x)=e^x
v'(x)=e^x
f'(x)=3e^x+3xe^x
3e^x+3xe^x=0<- to find CNS
3e^x(1+x)=0 e^ ANYTHING IS NEVER ZERO
1+x=0
x=-1 CN
f'(-2)=3e^-2(1-2)
f'(0)=3e^0(1-0) --> Positive!
f is increasing on interval (-1,@@)
and decreasing on (-@@,-1)
would increase product rule but we are smart and we realize hey, this is the derivative of 3xe^x
f''(x)=3e^x+(3e^x+3xe^x)
6e^x+3xe^x
3e^x(2+x)=0
x=-2
f''(-3)=3e^-3(2-3) gives NEGATIVE
f''(0)=3e^0(2-0)>0
f is concave up on (-2,@@)
down on (-@@,-2)
only way we know when it goes up to down is at inflection points
149 at drug test lab ate one pack of old (hard) swedish fish
ray expected to call in 45 minutes or an hour
ray never calls, tells me to fuck off when I call him> NO DRUG TEST TAKEN
Must find rent.
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