Summary Calculus 1 - Applications of Derivatives

Chapter 3 -

Applications of Derivatives


Related Rates


Ex two Car A the west of Car B and starts
driving to at 35 mph at the Car B starts at
.
cars start out 500 miles apart . is to the east same time
driving

50 mph After of what rate ?
. 3 hours
driving at is the distance between the two cars
changing
? Is it
increasing or
decreasing
500 miles

Car A carbinitial after 3 hours driving time : x= 500-35(3) = 395 50(3) = 150 we can use
pythagorean theorem to find
y
=

X


2 Y
Distance
carB 2 = 395" + 150 = 178525 -z = N8525 = 422 5222 determine 2 given that x = -35 and y = 50 (a rate is
negative if the
by
.

Driven
Car A




quantity is
decreasing) use implicit differentiation and pythagorean theorem 2 = x +
y - 222' = 2xx +
zyy

Canc a 2 from
everything and
plug in known quantities 2'(422 5222) .
=
(395)(-35) + (150)(50) -
z 6-14 9696
= .

decreasing at a rate of 14 .
969m


Critical Points

·


f(x) if f) and if either of the f'CC = f'cl doesn't
we
say that X = C is a critical point of the function exists
following are true 0 or exist



*
Ex . determine all the critical points for the function : f(x) = 6x + 33x" -30x3 + 100




f'(x) 30x" = + 132x 90x2 -
= 6x (5x + 22 -
15) = 6x (5x -
3)(X + 5) the only critical points will bethose values of X which make the derivative zero




6x2 (5x 3)(x + 5) - = 0 solu x= -
5 ,
x= 0 , x = 3/5


25-t g = j(t) =
0
a .
g(t) = (2t -
1 = some to and


b . R(w) == RSW) =+)
=-
40 -1 the derivative will not exist if there is division by Zero in the denominator solve : w w -
b = (w -
3)(w + 2) = 0
wa -
w
-
6 (w2 - w = 6)2 (wa - w - 6)


The derivative will not exist at W = 3 and w= 2 the numerator donsnt factor , use the quadratic formula
we
42 2
o



= -715E we have I critical points w = -7 + 55 and w = -7 - 52



2 f(x) = x (n(3x) + 6 f'(x) = 2x(n(3x) + x2 () =
2x(n(3x) + x = X(2(n(3x) + 1) this derivative will not exist if X is a negative number or if X =o




the derivative will
only be Zero if [In(3x) + 1 =0 In(3x) = -Ya sol
by exponentiating both sides e
In 3x
= e
-
/2
= 3x = e
-
"(2
X= Ye =
* * * *
d .
f(x) = xc f'(x) = c + xc (2x) = e (1 + 2x) this function will never be zero for any real value of X therefore this function will not have
any critical points



Minimum and Maximum values




I say that f(x) has absolute at if f(x) & fic) for domain
. We an maximum X= C
every in the we are
working on




z We
say that f(x) has a relative (or local) maximum at X= c if f(x) f(c) for x In some open interval around X =C
.


every


3 We
say that f(x)
has an absolute at X C if f(x) ]fCC) for x the domain we are working on
every in
.
minimum =




Y we
say that f(x) has relative local at X C if f(x) f() for some
open interval around X= c
or
every
a minimum = in




Note that when we "open interval around c" I
that will be contained somewhere inside the interval and will not be either of the endpoints
say
an X= we mean




·
we
collectively call the max and min points of a function the extrema : Relative extrema and absolute extrema



Ex .
identify the absolute and relative extrema for f(x) = x on 5-1 2] ,




4 -
·




3- looking at the
graph there is a relative and absolute min . of zero at X = O and an absolute max of 4 at x= .
2 Note that X I is not a relative


2 -




max . Since it is at the endpoint of the interval . this function doesn't hav
any relative maximums

·
-




- !

,Extreme Valve theorem



·

Suppose that f(x) is continuous on the interval [a b] then there
, are two numbers a c ,
d = b so that f(c) is an absolute maximum function and


fid) is an absolute minimum for the function

I

Ex .
f(x) = T on [-1 , 1]

25-

function function function does
at X = O
approaching infinity absolut
The is not continuous as we more in towards the is . So , the not have an
20 -




15-

maximum . Note that it does have an absolute minimum however the absolute min .
Occurs twice at X = 1 and X = /
10 -




-
5


I
- 1 '
Fermat's theorem


If
·


f(x) has a relative extrema at X = C and f'CC) exists then X= C is a critical point of f(x) . In fact ,
it will be a critical point such that f'() = 0




say that f'()
that
assuming that f'()
·
Note we can =O because we are also exists



Ex . f(x) = x3


we s a that this function has relative at X = 0 So Fermats theorem Should be critical f'(x) 2x and X = 0 critical point
a
according to X O point
minimum . = a = is a




* Fermat's theorem only works for Critical points in which f' = 0 .
this does not mean that relative extrema won't occur at critical points where the derivative DNE




Finding Absolut Extrema of f(x) on [a b] .




O .

Verify that the function is continuous on the interval [a , b]



I . Find all critical points of f(x) that are in the interval [a , b] Since we are
only interested in what the function is
doing in this interval ,
we don't care about critical points




that fall outside the interval



2 . Evaluate the function at the critical point found in stepl and the endpoints



3 .

Identify the absolute extrema



Ex .
g(t) = 2t + 3t -
12 + + 4 on 2 -
4 ,
2]



polynomial Continuous g(t) Ge 67-12 6/t + 2)(t-1) critical points : E 2 t 1 evaluate the function at the critical points and endpoints
everywhere +
= = = =- =
,




g) -2) = 24
g( 4) = 28 g() = 3
g(z) 8 = absolute max of 24 at t = -2 absolute min Of 28 at t = -4
- - -
.
,




Ex. Suppose that the amount of
money in a bank account aftert years is A(t) =
2000 - 10th
determine the min and max
during the first 10
years .
20 , 10] Alt) = +
10e
5-5) = im derivate exists
everywhe and fre



exponential is never
zero- =
0- = 4 -t = 2 I2 are both critical points but X= 2 is the
only one i n the interval ACO) = 2000 All = 199 66
.




ALIO) = 1999 . 44 the max is 2000 at t=o and the min is 199 66 . at t = 2




Shape of a Graph . PtI



1 . Given
any X , and X2 from an interval 1 with X . LX2 if f(x) Lf(x2) then f(x) is
increasing on



2 . Given any X , and xz from an intervall with X .< X2 If f(x) > f(x2) then f(x) is
decreasing on