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)

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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)

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do you know how to do this? @IrishBoy123
h'(1) = f'(1)-g'(1) u must find f'(x) and g'(x)

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where are you stuck?
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
h(x) = f(x)-(x) =\[x^{2}-x-\frac{ 1 }{ x }\] \[=2x-1-\frac{ 1 }{ x^{2} }\] Is this correct?
not quite last term?
the last term is wrong?
\[-\frac{ 1 }{ x } = - x^{-1}\]
differentiate that again
\[2x-1-\frac{ 1 }{x^{-2} }\]
2x-1-(-1/x^2) = f'(x)-g'(x)
@amy0799 its like my answer see it
\[\frac{d}{dx} (\frac{1}{x}) = \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}\] the sign changes :p
y wouldn't it be x^-2?
\[2x-1-(-1/x^2) = 2x-1+1/x^2\]
x^-2 = 1/x^2 u are right but we dont need it @amy0799
you had it right with our first go except you got the sign wrong on last term
me ? @IrishBoy123 u spoke with me
lol
no @dinamix :p i was addressing @amy0799
h(x)=f(x)g(x) \[=x^{2}-x(-\frac{ 1 }{ x^{2} })+\frac{ 1 }{ x }(2x-1)\] is this right so far?
@amy0799 yup this
\[\frac{ -x-1 }{ x }+\frac{ 2x-1 }{ x }\]
would this simplify to \[\frac{ -x-2 }{ x }?\]
-x+1/x @amy0799
not -x-1/x
i'm afraid i am not following what you have done so i shall leave it to you and @dinamix
ok @IrishBoy123 i help him ty anyway for coming
-1+(-1)=-2 I dont understand how you got -1
\[(x^2-x)(-1/x^2) = -1+\frac{ 1 }{ x } = \frac{ -x+1 }{ x}\]
u made mistake @amy0799 u find it ?
f(x)g(x) = 1 must find it like @amy0799 so check and tell me
\[(x^{2}-x)-\frac{ 1 }{ x ^{2} }=-\frac{ x ^{2} -x+1}{ x ^{2} }\]
is this how you set it up?
\[\frac{ -x^2+x }{ x^2 } \]
its like that @amy0799
oh ok
so what u find f(x)g(x) = i want your answer
h(x)=f(x)/g(x) \[=\frac{ \frac{ 1 }{ x(2x-1)-(x ^{2}-x)-\frac{ 1 }{ x ^{2} } } }{ (\frac{ 1 }{ x })^{2} }\]
is this setup right?
i messed up, it's 1/x
u make some mistake
\[\frac{ \ ( \frac{ 1 }{ x})( 2x-1) -[\frac{ -1 }{ x^2} (x^2-x)]} { \frac{ -1 }{ x^2 } }\]
its not - 1 only 1
in demonst
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?
\[\frac{ (2x-1+x-1)x^2 }{ x }\]
we calcul h'(x)
by using f'(x) and g'(x) thats only @IrishBoy123
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.
if you want to collate your solutions, i am happy to come back later and check them @freckles
we have 2 method @amy0799
\[\frac{ (3x-2)x ^{2} }{ x }\] can this simplify further?
yup
h'(x) for the last one
h'(x) for function f(x)/g(x) = (3x^2-2x)
i mean h'(x) =(3x^2-2x) not f(x)/g(x) did u understand
all this time we calcul h'(x) not h(x) @amy0799
did u understand mate what we are do
woudn't it be \[\frac{ (3x ^{3}-2x ^{2}) }{ x }?\]
x^2/x = x right @amy0799 and its same
and u have devide by x
oh ok. so 3x^2-2x is the answer?
yup
the question also asked arrange the values of h'(1) from smallest to largest. do i plug in 1 to x?
and this h'(x) for function f(x)/g(x) , u understand what was do
we find all h'(x) for this fuction f(x)-g(x) and f(x)g(x) and f(x) /g(x)
right ?
right, so what does arrange the values of h'(1) from smallest to largest mean?
h'(1) = f'(1)-g'(1) = 2(1)-1+1(1^2) = ?
thats only
=2 right?
1- h'(1) = 2 yup
for f(x)/g(x) 3(1)^2-2(1)=1
yup this
and for fuction f(x)g(x) ; h'(1) = ?
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
not f(x)g(x) =1 its f'(x)g'(x) = 1
why u make this mistake ?@amy0799
what mistake?
look up what i write u
not f(x)g(x)= 1 its f'(x)g'(x) = 1
oh i forgot to put in the '
are u sure u understand now .@amy0799
yes so the order goes f'(x)g'(x)=1 f'(x)/g'(x)=1 f'(x)-g'(x)=2 ?
f'(1)g'(1)=1 f'(1)/g'(1) =1 f'(1)-g'(1)=2 not like u said
oh ok thank you. can you help me with another problem?
ok but fast
if f(x)=x^3+2x^2-5x-1 find \[\lim_{h \rightarrow 0}\frac{ f'(-3+h)-f'(-3) }{ h }\]
\[\lim_{h \rightarrow 0}\frac{ f(-3+h)-f(-3) }{ h}\]
@amy0799 not like u said its not f'(-3+h) -f'(-3) look good how i write it
that's how my question was written
made mistake can u check plz cuz i didnt see it my life like what u write ?
i am sure my write is true @amy0799
you could be right, im not sure, can you help me solve it plz
f(-3+h) = (-3+h)^3+2(-3+h)^2-5(-3+h)-1
the same thing for f(-3) = @amy0799
should i find the derivative of f(x)=x^3+2x^2-5x-1 first?
\[\lim_{h \rightarrow 0} \frac{ f(-3+h)-f(-3) }{ h } = f'(-3)\]
@amy0799 u can find it i cant help u more u should try
cuz i will go to sleep its 1:00 Am
ok thank you for all your help!!

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