Question

Find the maximum and minimum values of the function on the given interval.
y = x^2 – 2x – 7, [-2, 0]

Answer 1

Find y' have you?

Answer 2

And actually we don't need calculus for this one

Answer 3

This is a parabola you can put this in vertex form

Answer 4

to determine the vertex which will be a min

Answer 5

since the parabola is concave up

Answer 6

But we do have endpoints

Answer 7

so we need to see if the x-coordinate of the vertex is btw -2 and 0

Answer 8

Which way do you prefer?
Cal or Algebra?

Answer 9

actually am doing a calculus based course,so i would like to know it in calculus :)

Answer 10

SO have you found y'?

Answer 11

that would simply be 2x-2

Answer 12

\[y'=(x^2)'-(2x)'-(7)'=2x-2=2(x-1)\]
Now set y'=0

Answer 13

and solve for x

Answer 14

Like this:
2(x-1)=0

Answer 15

But guess what happens?

Answer 16

x=1 is not btw -2 and 0
so we do not care about x=1
We only have to look the endpoints

Answer 17

ok

Answer 18

So evaluate y for x=0
Then evaluate y for x=-2

Answer 19

Which is higher output and which is the lower output?

Answer 20

how do i evaluate?

Answer 21

\[y|_{x=0}=0^2-2(0)-7=-7\]
\[y|_{x=-2}=(-2)^2-2(-2)-7=4+4-7=8-7=1\]

Answer 22

See I evaluated y for x=0 and x=-2

Answer 23

Now 1 is higher than -7
So 1 is the max value
and
-7 is the min value
on the interval [-2,0]

Answer 24

if so why do i need to derive the calculus? if i can just substitute into the original function to get the max and the mini?

Answer 25

Because we needed to figure out if we had any critical numbers btw the endpoints

Answer 26

For example if the interval was [0,2] we would have gotten something differently ....

Answer 27

We had x=1 is the critical number
1 is btw 0 and 2
so we have to check both endpoints and the critical number for this one

Answer 28

wat of in cases like this
Find the maximum and minimum values of the function on the given interval.
y = (3 - x)/(x^2 + 9 x), text( ) text( ) [3, 10]

Answer 29

Find y'
Set y'=0
Also find where y' DNE
These will give you your critical numbers
Make sure you only look at the critical numbers btw 3 and 10

Answer 30

u can just go on with ur example,if i learn i can just solve this one i posted and the rest.
as u were saying if 1 happens to be with the given intervals,how will i get the max and the mini from there?

Answer 31

\[y'=\frac{-1(x^2+9x)-(3-x)(2x+9)}{(x^2+9x)^2}=\frac{-x^2-9x-(6x+27-2x^2-9x)}{(x^2+9x)^2}\]
\[=\frac{-x^2-9x-(-2x^2-3x+27)}{(x^2+9x)^2}\]
=\[\frac{-x^2-9x+2x^2+3x-27}{(x^2+9x)^2}=\frac{x^2-6x-27}{(x^2+9x)^2}\]
So where is y'=0 and where does y' DNE ?

Answer 32

Oh you want to do the one I was doing lol sorry

Answer 33

y does not exist when x=0

Answer 34

dont worry just go on with this

Answer 35

dis will give me a better understanding

Answer 36

Ok so we had \[y=x^2-2x-7, [0,2]\]
So we find y' just like we did before
y'=2x-2=2(x-1)
Set y'=0 and we get x=1
So 1 is btw the endpoints so we have to check all three of these x-values

Answer 37

\[y|_{x=0}=0^2-2(0)-7=-7\]
\[y|_{x=1}=1^2-2(1)-7=1-2-7=1-9=-8\]
\[y|_{x=2}=2^2-2(2)-7=-7\]
-8 is smallest so -8 is the min value
-7 is highest so -7 is the max value
on [0,2]

Answer 38

Ok so if we were looking at this new problem you put up

Answer 39

yeah

Answer 40

\[y'=\frac{x^2-6x-27}{(x^2+9x)^2}\]

Answer 41

To find where y'=0
Set top=0 and solve for x
\[x^2-6x-27=0 => (x-9)(x+3)=0 \]
x=? or x=?
-----------------------------
To find where y' DNE
Set bottom=0 and solve for x
\[(x^2+9x)^2=0 => x^2+9x=0 =>x(x+9)=0\]
x=? or x=?
-------------------------------

Answer 42

What are the critical numbers?

Answer 43

-9 and 0?

Answer 44

and ...

Answer 45

9 and -3 right?

Answer 46

now we only want to look at the critical numbers on the interval given

Answer 47

So which are those?

Answer 48

why -3?

Answer 49

-3+3=0

Answer 50

You did solve both equations right?

Answer 51

i solved for x(x+9)=0

Answer 52

What about the other equation I had there to solve?

Answer 53

Did you read everything I typed?

Answer 54

oh! now i see the equation,am sorry :)

Answer 55

u can go on

Answer 56

ok so the interval was [3,10] right?

Answer 57

yep

Answer 58

Now which of the critical numbers we found are btw 3 and 10?

Answer 59

9

Answer 60

ok so we need to evaluate y for x=3
evaluate y for x=9
evaluate y for x=10

Answer 61

\[y=\frac{3-x}{x^2+9x}\]
\[y|_{x=3}=\frac{3-3}{3^2+9(3)}=0\]
\[y|_{x=9}=\frac{3-9}{9^2+9(9)}=\frac{-6}{2(81)}=\frac{-3}{81}\]
\[y|_{x=10}=\frac{3-10}{10^2+9(10)}=\frac{-7}{190}\]

Answer 62

Which is the lowest output?

Answer 63

-7/190

Answer 64

so that's my mini right?

Answer 65

-3/81 is smallest so -3/81 is the min value
0 is highest so 0 is the max value

Answer 66

ok,so i can use this procedure to solve the rest?

Answer 67

what do you mean?

Answer 68

like a question like this y = (sqrt(5 + x^2)) - 4 x, [5, 6]
i can use ur procedure to determine the max and the mini?

Answer 69

of course
y'=0 and y' DNE
Use all critical numbers btw the endpoints given and the endpoints and plug them into y
see what the lowest output (min value) is and then see what the highest output (max value) is

Answer 70

ok thanks alot for your help :)

Answer 71

By the way when finding values for y' DNE make sure that y exists for those x values

Answer 72

oh ok sure :)

Answer 73

@myininaya i have a problem with the remaining two quesrtions

Answer 74

two?

Answer 75

yeah in this situation,i set my function=0 and got
x-2((4+x^2)^1/2)

Answer 76

is my cn just 0 in this case?

Answer 77

I don't understand...Are these the same type of questions as before?

Answer 78

yeah

Answer 79

So what is y?

Answer 80

y = (sqrt(4 + x^2)) - 2 x, [4, 5]

Answer 81

\[y'=\frac{x}{\sqrt{4+x^2}}-2\]Is that what you got for y'?

Answer 82

yeah

Answer 83

the i set it to 0

Answer 84

Yes

Answer 85

\[\frac{x-2\sqrt{4+x^2}}{\sqrt{4+x^2}}=0\]

Answer 86

got x-2(4+x^2)^1/2

Answer 87

yea

Answer 88

\[x-2 \sqrt{4+x^2}=0 \]
\[x=2 \sqrt{4+x^2} => x^2=4(4+x^2) \text{ squaring both sides}\]

Answer 89

ok,so my cn will be?

Answer 90

\[x^2=16+4x^2 =. x^2-4x^2=16 => -3x^2=16 => x^2=\frac{16}{-3}\]
But this will give us imaginary x
So we don't have any x-values that satisfy y'=0
So now look for when y' DNE

Answer 91

But again no such values exist

Answer 92

So last thing to do is just plug in endpoints

Answer 93

which i s [4 and 5] ?

Answer 94

\[y|_{x=4} =\sqrt{4+4^2}-2(4) \]
\[y|_{x=5}=\sqrt{4+5^2}-2(5)\]

Answer 95

so that will be my max and mini?

Answer 96

the lowest ouput=min value
the highest output=max value

Answer 97

\[y=5e^x-e^{3x}\] ?

Answer 98

yeah exactly {-1,1}