Question

If f(x)=x^2-x and g(x)=1/x, and h(x) is defined with the expressions below, arrange the values of h'(1) from smallest to largest.
1. h(x) = f(x)-g(x)
2. h(x) = f(x)g(x)
3. h(x) = f(x)/g(x)

Answer 1

do you know how to do this? @IrishBoy123

Answer 2

h'(1) = f'(1)-g'(1) u must find f'(x) and g'(x)

Answer 3

@amy0799

Answer 4

where are you stuck?

Answer 5

for 2- h'(1) =f'(1)g(1)+g'(1)f(1)
3- h'(1) =( f'(1) g(1) - f(1)g'(1) )/ (g(1))^2 this the answer @IrishBoy123 right lol

Answer 6

h(x) = f(x)-(x)
=\[x^{2}-x-\frac{ 1 }{ x }\]
\[=2x-1-\frac{ 1 }{ x^{2} }\]
Is this correct?

Answer 7

not quite
last term?

Answer 8

the last term is wrong?

Answer 9

\[-\frac{ 1 }{ x } = - x^{-1}\]

Answer 10

differentiate that again

Answer 11

\[2x-1-\frac{ 1 }{x^{-2} }\]

Answer 12

2x-1-(-1/x^2) = f'(x)-g'(x)

Answer 13

@amy0799 its like my answer see it

Answer 14

\[\frac{d}{dx} (\frac{1}{x}) = \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}\]

the sign changes :p

Answer 15

y wouldn't it be x^-2?

Answer 16

\[2x-1-(-1/x^2) = 2x-1+1/x^2\]

Answer 17

x^-2 = 1/x^2 u are right but we dont need it @amy0799

Answer 18

you had it right with our first go except you got the sign wrong on last term

Answer 19

me ? @IrishBoy123 u spoke with me

Answer 20

lol

Answer 21

no @dinamix :p

i was addressing @amy0799

Answer 22

h(x)=f(x)g(x)
\[=x^{2}-x(-\frac{ 1 }{ x^{2} })+\frac{ 1 }{ x }(2x-1)\]
is this right so far?

Answer 23

@amy0799 yup this

Answer 24

\[\frac{ -x-1 }{ x }+\frac{ 2x-1 }{ x }\]

Answer 25

would this simplify to \[\frac{ -x-2 }{ x }?\]

Answer 26

-x+1/x @amy0799

Answer 27

not -x-1/x

Answer 28

i'm afraid i am not following what you have done so i shall leave it to you and @dinamix

Answer 29

ok @IrishBoy123 i help him ty anyway for coming

Answer 30

-1+(-1)=-2
I dont understand how you got -1

Answer 31

\[(x^2-x)(-1/x^2) = -1+\frac{ 1 }{ x } = \frac{ -x+1 }{ x}\]

Answer 32

u made mistake @amy0799 u find it ?

Answer 33

f(x)g(x) = 1 must find it like @amy0799 so check and tell me

Answer 34

\[(x^{2}-x)-\frac{ 1 }{ x ^{2} }=-\frac{ x ^{2} -x+1}{ x ^{2} }\]

Answer 35

is this how you set it up?

Answer 36

\[\frac{ -x^2+x }{ x^2 } \]

Answer 37

its like that @amy0799

Answer 38

oh ok

Answer 39

so what u find f(x)g(x) = i want your answer

Answer 40

h(x)=f(x)/g(x)
\[=\frac{ \frac{ 1 }{ x(2x-1)-(x ^{2}-x)-\frac{ 1 }{ x ^{2} } } }{ (\frac{ 1 }{ x })^{2} }\]

Answer 41

is this setup right?

Answer 42

i messed up, it's 1/x

Answer 43

u make some mistake

Answer 44

\[\frac{ \ ( \frac{ 1 }{ x})( 2x-1) -[\frac{ -1 }{ x^2} (x^2-x)]} { \frac{ -1 }{ x^2 } }\]

Answer 45

@amy0799

Answer 46

its not - 1 only 1

Answer 47

in demonst

Answer 48

where did this come from?!?!?!

https://gyazo.com/b501b2ad7a4061eab75c1cf7645b4f67

\[h(x) = \frac{f(x)}{g(x)} = \frac{x^2 - x}{\frac{1}{x}} = x(x^2 - x)\]

now, expand and differentiate. that's all there is to it

and did we ever come up with an answer to pt 1?

Answer 49

\[\frac{ (2x-1+x-1)x^2 }{ x }\]

Answer 50

we calcul h'(x)

Answer 51

by using f'(x) and g'(x) thats only @IrishBoy123

Answer 52

and what exactly is this all about?!?!?!?!

https://gyazo.com/d2c0f8f760ddd3e17b317893a135989d

\[h(x) = f(x)g(x) = (x^2 - x). \left(\frac{1}{x}\right) = x - 1\]
\[h'(x) = 1\]

done.

Answer 53

if you want to collate your solutions, i am happy to come back later and check them

@freckles

Answer 54

we have 2 method @amy0799

Answer 55

\[\frac{ (3x-2)x ^{2} }{ x }\]
can this simplify further?

Answer 56

yup

Answer 57

h'(x) for the last one

Answer 58

h'(x) for function f(x)/g(x) = (3x^2-2x)

Answer 59

i mean h'(x) =(3x^2-2x) not f(x)/g(x) did u understand

Answer 60

all this time we calcul h'(x) not h(x) @amy0799

Answer 61

did u understand mate what we are do

Answer 62

woudn't it be \[\frac{ (3x ^{3}-2x ^{2}) }{ x }?\]

Answer 63

x^2/x = x right @amy0799 and its same

Answer 64

and u have devide by x

Answer 65

oh ok. so 3x^2-2x is the answer?

Answer 66

yup

Answer 67

the question also asked arrange the values of h'(1) from smallest to largest. do i plug in 1 to x?

Answer 68

and this h'(x) for function f(x)/g(x) , u understand what was do

Answer 69

we find all h'(x) for this fuction f(x)-g(x) and f(x)g(x) and f(x) /g(x)

Answer 70

right ?

Answer 71

right, so what does arrange the values of h'(1) from smallest to largest mean?

Answer 72

h'(1) = f'(1)-g'(1) = 2(1)-1+1(1^2) = ?

Answer 73

thats only

Answer 74

=2
right?

Answer 75

1- h'(1) = 2 yup

Answer 76

for f(x)/g(x)
3(1)^2-2(1)=1

Answer 77

yup this

Answer 78

and for fuction f(x)g(x) ; h'(1) = ?

Answer 79

so from smallest to largest it would be f(x)g(x), f(x)/g(x), f(x)-g(x)
f(x)g(x)=1

Answer 80

not f(x)g(x) =1 its f'(x)g'(x) = 1

Answer 81

@amy0799

Answer 82

why u make this mistake ?@amy0799

Answer 83

what mistake?

Answer 84

look up what i write u

Answer 85

not f(x)g(x)= 1 its f'(x)g'(x) = 1

Answer 86

oh i forgot to put in the '

Answer 87

are u sure u understand now .@amy0799

Answer 88

yes so the order goes
f'(x)g'(x)=1
f'(x)/g'(x)=1
f'(x)-g'(x)=2
?

Answer 89

f'(1)g'(1)=1
f'(1)/g'(1) =1
f'(1)-g'(1)=2 not like u said

Answer 90

@amy0799

Answer 91

oh ok thank you. can you help me with another problem?

Answer 92

ok but fast

Answer 93

if f(x)=x^3+2x^2-5x-1
find
\[\lim_{h \rightarrow 0}\frac{ f'(-3+h)-f'(-3) }{ h }\]

Answer 94

\[\lim_{h \rightarrow 0}\frac{ f(-3+h)-f(-3) }{ h}\]

Answer 95

@amy0799 not like u said its not f'(-3+h) -f'(-3) look good how i write it