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egoroff_w [7]
3 years ago
14

Domain of y= 6x+13/x^2-4x-12

Mathematics
1 answer:
borishaifa [10]3 years ago
5 0
Domain is the set of numbers you can subsitute for the input (in this case x) and for it to be allowed (dividing by zero is not allowed so if you have y=6/x then x is not allowed to be zero, it can be anything else though)

ok, so from experience the x^2-4x-12 part is supposed to be all under the parenthaseeese and if so, go to part AAAAAAAAAAAAA
if it isn't and only x^2 is below the fraction, then go to BBBBBBBB


AAAAAAAAA
y= \frac{6x+13}{x^{2}-4x-12}
so the domain must be valid
one rule that we have is that you cannot divide by zero
therefor we find what numbers make the denomenator zero and don't allow them
x^2-4x-12=0
factor
(x-6)(x+2)=0
x-6=0
x=6

x+2=0
x=-2

so since -2 and 6 make the thingummy zero, we say that

domain={all real numbers except for -2 and 6}




BBBBBBBBBBBBBBB
y= \frac{6x-13}{x^{2}} -4x-12

so remember that you cannot divide by zero
therefor you have to make sure the deomenator is not equal to zero by finding those values and restricting them from the solution set
so
x^2=0
squaer root both sides
x=0
domain={all real numbers except for 0}





so if the x^2-4x-12 is under the dividing line, then the answer is domain={all real numbers except for -2 and 6}

if the x^2 only is under the dividing line then the answer is domain={all real numbers except for 0}
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Use the distributive property to express 30 + 75
OleMash [197]
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MariettaO [177]

Answer:

sorry to answer late but..

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Can anyone help me integrate :
worty [1.4K]
Rewrite the second factor in the numerator as

2x^2+6x+1=2(x+2)^2-2(x+2)-3

Then in the entire integrand, set x+2=\sqrt3\sec t, so that \mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt. The integral is then equivalent to

\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt
=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt

Note that by letting x+2=\sqrt3\sec t, we are enforcing an invertible substitution which would make it so that t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3} requires 0\le t or \dfrac\pi2. However, \tan t is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which \tan t>0 so that |\tan t|=\tan t. This allows us to write

=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt
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We can show pretty easily that

\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C
\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C
\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C
\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C

which means the integral above becomes

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Back-substituting to get this in terms of x is a bit of a nightmare, but you'll find that, since t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}, we get

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\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}
\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}

etc.
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