Question

1. Find the absolute maximum and minimum values of the function f(x)=12+4x-x^2 in the interval [0,5]

Answer 1

find the derivative,
and, set it equal to 0

Answer 2

that gives u critical points where max/min values can occur

Answer 3

yes, I did that.

so it'll only be x=2?

Answer 4

x = 2 is one possibility where max/min value can occur

Answer 5

other possibilities are x = 0 and x = 5

Answer 6

yes yes :)

I'm not sure but I got 12, 7 and 16 :)

Answer 7

so, here is the thing :
Absolute maximum = Maximum of {f(2), f(0), f(5)}
Absolute minimum = Minimum of {f(2), f(0), f(5)}

Answer 8

I thought there can be only a maximum of 4 values to choose from? :)

Answer 9

so :
Absolute maximum = Maximum of {16, 12, 7} = 16
occurs when x = 2

Answer 10

Absolute manimum = Minimum of {16, 12, 7} = 7
occurs when x = 5

Answer 11

since you're given BOUNDARY, both max and min will be there.

Answer 12

if you're not given any boundary, then oly max will be there for the given function

Answer 13

ohh :)

Answer 14

can you please help me with 2 more? it's harder :(

Answer 15

so to find absolute max/mins : u need to check both 'local max/mins' and the 'boundary values'

Answer 16

sure :)

Answer 17

thankyou so much

2. Find the absolute maximum and minimum values of the function \[g(x)=(x^{2}-1)^{3}\] in the interval [-1,2]

Answer 18

step1 : take the derivative, set it equal to 0, and solve x

Answer 19

okay, Im going to try it :)

Answer 20

chain rule? :O

Answer 21

yess

Answer 22

I'm now on \[6x(x ^{2}-1)^{2}=0\] and I really don't know what to do with it :"(

Answer 23

solve x

Answer 24

use zero-product-property

Answer 25

oh okay, I think I get it

Answer 26

is it x=0,1 ?

Answer 27

u left one value :)

Answer 28

\(\large 6x(x^2-1) = 0 \)
\(\large \implies x = 0 \) or \(\large \implies x^2-1 = 0 \)

Answer 29

\(\large x^2-1 = 0 \implies x^2 = 1 \implies x = \pm 1\)

Answer 30

so the local max/mins are : x = 0, 1, -1

Answer 31

okay ?

Answer 32

oh okay :)

Answer 33

step2 : evaluate the given function at 1) local max/mins and 2) boundary

Answer 34

I'm going to solve for it now :)

Answer 35

okie

Answer 36

max:9 min:0?

Answer 37

basically u need to evaluate below :
g(0)
g(-1)
g(1)
g(2)

Answer 38

let me check..

Answer 39

\(\large g(x) = (x^2-1)^3\)

g(0) = -1
g(-1) = 0
g(1) = 0
g(2) = 27

Answer 40

so, minimum value = -1, occurs when x = 0

Answer 41

oh no. haha

Answer 42

maximum value = 27, occurs when x = 2

Answer 43

sorry. my bad. I wrote here (3)^3, thought it was 3 x 3

Answer 44

haha i guessed that :)

Answer 45

graph looks like that,
it may be exciting to see that the graph has minimum value at x = 0 ... :)

Answer 46

oh cool graph. haha

sorry, another careless mistake. I wrote -1 here and I encircled it but I didn't saw it :(

Answer 47

last item number :)

I find this really difficult because it involves trigonometry :(

Answer 48

oh lets see -.-

Answer 49

3. Find the absolute maximum and minimum values of the function h(x)=2cost + sin2t in the interval\[[0,\pi/2]\]

Answer 50

step1 : take the derivative, set it equal to 0, and solve x

Answer 51

*or t, or watever the variable is

Answer 52

uhm, how do you find the derivative when its sin2t again? haha I completely forgot about it sorry

Answer 53

chain rule to the rescue !

Answer 54

ohh haha

Answer 55

\(\large \frac{d}{dt}[\sin(2t)] = \cos(2t) * \frac{d}{dt}(2t) = \cos(2t) * 2 = 2\cos(2t)\)

Answer 56

first, find the derivative of h(t) and set it equal to 0

later we can see how to solve it :)

Answer 57

okay :) i'll try my best

Answer 58

but ...

Answer 59

good, give it a try :)

Answer 60

\(\large h(t) = 2\cos(t) + \sin(2t)\)

\(\large h'(t) = -2\sin(t) + 2\cos(2t)\)

Answer 61

set it equal to 0

Answer 62

\(\large h'(t) = 0\)

\(\large -2\sin(t) + 2\cos(2t) = 0 \)

\(\large -\sin(t) + \cos(2t) = 0 \)

\(\large -\sin(t) + 1-2\sin^2t= 0 \)

\(\large 2\sin^2t + \sin t -1 = 0 \)

Answer 63

but how? it has sin and cos? :"(

Answer 64

we got rid of cos's already :)
go thru it and see if it looks okay... let me knw if something doesnt make sense ..

Answer 65

we still need to solve to \(t\) ...

Answer 66

is 1 -2sin2t an identity I should know about? :O

Answer 67

good question :)

yes u need to memorize below trig identities :
\(\large \cos 2t = 1-2\sin^2 t\)

\(\large \cos 2t = 2\cos^2 t-1\)

Answer 68

the 2nd one, I think that should be sine? :)

Answer 69

nope. its cos oly..

Answer 70

really? are there any more identities? :)

Answer 71

one more important one is :
\(\large \sin^2 t + \cos^2t = 1\)

Answer 72

another important one is :
\(\large \sin 2t = 2 \sin t \cos t\)

Answer 73

those four identities will do the job most of the time...

Answer 74

oh okaay thankyou so much and now back to solving :)

Answer 75

yeah where are we..

Answer 76

\(\large 2\sin^2t + \sin t -1 = 0 \)

we are at solving this equation for \(t\)

Answer 77

huh? not sin(t)? O_O

Answer 78

nope, we want to find \(t\) that gives the min/max values

Answer 79

ohh

Answer 80

... for the given function h(t)

Answer 81

\(\large 2\sin^2t + \sin t -1 = 0 \)

if u replace \(\sin t\) wid \(x\), clearly its a quadratic equation in x's right ?

Answer 82

yeah :)

Answer 83

so, can u factor it ha ? :)

Answer 84

(2x+1)(x-1)

Answer 85

\[(2sint +1 ) (sint -1)\]

Answer 86

is it correct? :)

Answer 87

partially correct...
^careful about the signs

Answer 88

\(\large 2\sin^2t + \sin t -1 = 0 \)

\(\large 2\sin^2t + 2\sin t - \sin t -1 = 0 \)

\(\large 2\sin t(\sin t + 1) - 1( \sin t + 1) = 0 \)

\(\large 2\sin t(\sin t + 1) - 1( \sin t + 1) = 0 \)

\(\large (2\sin t - 1)( \sin t + 1) = 0 \)

Answer 89

okay ?

Answer 90

sorry :)

Answer 91

np :) solve t

Answer 92

\[\sin(t)=\frac{ 1 }{ 2 },-1\]

Answer 93

perfect ! solve t

Answer 94

\(\large (2\sin t - 1)( \sin t + 1) = 0 \)

\(\large \implies 2\sin t - 1 = 0\) or \(\large \sin t + 1 = 0\)
\(\large \implies \sin t = \frac{1}{2}\) or \(\large \sin t = -1\)
\(\large \implies t = \frac{\pi}{3}\)

Answer 95

so oly one value we got for \(t\) after so much work -.-

Answer 96

btw, we had to discard -1 cuz that gives t = pi,
but the question is asking us to find in the boundary (0, pi/2)