amy0799
  • amy0799
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amy0799
  • amy0799
do you know how to do this? @IrishBoy123
dinamix
  • dinamix
h'(1) = f'(1)-g'(1) u must find f'(x) and g'(x)
dinamix
  • dinamix
@amy0799

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More answers

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

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