katherinesmith
  • katherinesmith
What is the approximate value of the function at x = 1? graph inside --->
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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katherinesmith
  • katherinesmith
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katherinesmith
  • katherinesmith
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
look at the graph locate x = 1 on the x axis and see where it lands on the curve

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More answers

jim_thompson5910
  • jim_thompson5910
the value of the function is equal to the y coordinate of the point at x = 1
katherinesmith
  • katherinesmith
that sentence just killed my brain cells
jim_thompson5910
  • jim_thompson5910
what's not making sense?
katherinesmith
  • katherinesmith
wait wait wait. so the curve lands on -0.75 correct? because once it touches that box that's where its closest to
jim_thompson5910
  • jim_thompson5910
yeah roughly, you are correct
jim_thompson5910
  • jim_thompson5910
when x = 1, y is y = -0.75 roughly
katherinesmith
  • katherinesmith
do you know how to find the range and domain of functions
jim_thompson5910
  • jim_thompson5910
yes I do
katherinesmith
  • katherinesmith
find the range of \[f (x) = \frac{ 2 }{ x - 3 } - 5\]
jim_thompson5910
  • jim_thompson5910
how would you find the horizontal asymptote of that
katherinesmith
  • katherinesmith
i have no idea... don't think i've learned that
jim_thompson5910
  • jim_thompson5910
oh wait, I didn't simplify first, hold on
jim_thompson5910
  • jim_thompson5910
\[f (x) = \frac{ 2 }{ x - 3 } - 5\] \[f (x) = \frac{ 2 }{ x - 3 } - 5\frac{x-3}{x-3}\] \[f (x) = \frac{ 2 }{ x - 3 } - \frac{5(x-3)}{x-3}\] \[f (x) = \frac{ 2 }{ x - 3 } - \frac{5x-15}{x-3}\] \[f (x) = \frac{ 2 - (5x-15) }{ x - 3 }\] \[f (x) = \frac{ 2 - 5x+15 }{ x - 3 }\] \[f (x) = \frac{ -5x + 17 }{ x - 3 }\] Since the degree of the numerator equals the degree of the denominator (they are both equal to 1), this means that the horizontal asymptote is y = -5/1 = -5
jim_thompson5910
  • jim_thompson5910
the range in interval notation is (-infinity, -5) U (-5, infinity)
katherinesmith
  • katherinesmith
i think i understand. now what about the domain of a different problem
katherinesmith
  • katherinesmith
\[f (x) = \frac{ 1 }{ x + 1 } - 4\]
jim_thompson5910
  • jim_thompson5910
hint: you cannot divide by zero
katherinesmith
  • katherinesmith
now that i know how to find the range i need to know how to find the domain.
jim_thompson5910
  • jim_thompson5910
if you cannot divide by zero, what does that mean for the domain?
katherinesmith
  • katherinesmith
no solution?
jim_thompson5910
  • jim_thompson5910
let's say the denominator was zero, so that would mean that x+1 = 0 what is x?
katherinesmith
  • katherinesmith
-1
jim_thompson5910
  • jim_thompson5910
therefore, if x = -1, then x+1 is zero
jim_thompson5910
  • jim_thompson5910
which means that you can plug in any value you want for x...BUT...x cannot be -1
jim_thompson5910
  • jim_thompson5910
x cannot be -1 to avoid division by zero everything else is just fine
katherinesmith
  • katherinesmith
so the domain is all real numbers except -1
jim_thompson5910
  • jim_thompson5910
exactly
katherinesmith
  • katherinesmith
thank you! alright next question,
katherinesmith
  • katherinesmith
\[\frac{ 2x - 5 }{ x - 2 } - 2 = \frac{ 3 }{ x + 2 }\]
jim_thompson5910
  • jim_thompson5910
\[\frac{ 2x - 5 }{ x - 2 } - 2 = \frac{ 3 }{ x + 2 }\] \[\frac{ 2x - 5 }{ x - 2 } - 2\frac{x-2}{x-2} = \frac{ 3 }{ x + 2 }\] \[\frac{ 2x - 5 }{ x - 2 } - \frac{2(x-2)}{x-2} = \frac{ 3 }{ x + 2 }\] \[\frac{ 2x - 5 }{ x - 2 } - \frac{2x-4}{x-2} = \frac{ 3 }{ x + 2 }\] \[\frac{ 2x - 5-(2x-4)}{x-2} = \frac{ 3 }{ x + 2 }\] \[\frac{ 2x - 5-2x+4}{x-2} = \frac{ 3 }{ x + 2 }\] I'll let you finish
katherinesmith
  • katherinesmith
@jim_thompson5910 i got x = 1
jim_thompson5910
  • jim_thompson5910
you are correct
katherinesmith
  • katherinesmith
thank god you literally saved my life
jim_thompson5910
  • jim_thompson5910
not really lol, but I'm glad to be of help
katherinesmith
  • katherinesmith
i do have another if you wanna stick around \[\frac{ x - 4 }{ 4 } + \frac{ x }{ 3 } = 6\]
katherinesmith
  • katherinesmith
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
I'll start you off \[\frac{ x - 4 }{ 4 } + \frac{ x }{ 3 } = 6\] \[\frac{ 3*(x - 4) }{ 3*4 } + \frac{ 4x }{ 4*3 } = 6\] \[\frac{ 3x - 12 }{ 12 } + \frac{ 4x }{ 12 } = 6\] \[\frac{ 3x - 12 + 4x }{ 12 } = 6\]
katherinesmith
  • katherinesmith
can you cancel -12 and 12 to get rid of your denominator?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
that is not a valid algebraic move
katherinesmith
  • katherinesmith
x = 12
jim_thompson5910
  • jim_thompson5910
good
katherinesmith
  • katherinesmith
now i have another domain question
katherinesmith
  • katherinesmith
\[f(x) = \frac{ 2 }{ x + 1 } - 3\]
jim_thompson5910
  • jim_thompson5910
tell me what you get
katherinesmith
  • katherinesmith
all real numbers except -1
jim_thompson5910
  • jim_thompson5910
you are correct
katherinesmith
  • katherinesmith
then what is the domain of \[f (x) = - \frac{ 3 }{ x - 4 } + 2\]
katherinesmith
  • katherinesmith
would it be all real numbers except 4?
jim_thompson5910
  • jim_thompson5910
very good
jim_thompson5910
  • jim_thompson5910
you got the hang of this domain stuff
katherinesmith
  • katherinesmith
yessss okay next one
katherinesmith
  • katherinesmith
\[\frac{ 3 }{ n - 1 } + \frac{ 2 }{ 3 } = \frac{ 5 }{ 3 }\]
jim_thompson5910
  • jim_thompson5910
\[\large \frac{ 3 }{ n - 1 } + \frac{ 2 }{ 3 } = \frac{ 5 }{ 3 }\] \[\large \frac{ 3*3 }{ 3(n - 1) } + \frac{ 2(n-1) }{ 3(n-1) } = \frac{ 5 }{ 3 }\] \[\large \frac{ 9 }{ 3(n - 1) } + \frac{ 2n-2 }{ 3(n-1) } = \frac{ 5 }{ 3 }\] \[\large \frac{ 9 + 2n-2 }{ 3(n-1) } = \frac{ 5 }{ 3 }\]
katherinesmith
  • katherinesmith
n = 4
jim_thompson5910
  • jim_thompson5910
correct
katherinesmith
  • katherinesmith
you are the bomb
katherinesmith
  • katherinesmith
i have 3 questions left
katherinesmith
  • katherinesmith
\[\frac{ 4 }{ n } - \frac{ 1 }{ 2 } = -\frac{ 6 }{ 3n }\]
jim_thompson5910
  • jim_thompson5910
\[\large \frac{ 4 }{ n } - \frac{ 1 }{ 2 } = -\frac{ 6 }{ 3n }\] \[\large \frac{ 2*4 }{ 2*n } - \frac{ n*1 }{ n*2 } = -\frac{ 6 }{ 3n }\] \[\large \frac{ 8 }{ 2n } - \frac{ n }{ 2n } = -\frac{ 6 }{ 3n }\] \[\large \frac{ 8 - n }{ 2n } = -\frac{ 6 }{ 3n }\]
katherinesmith
  • katherinesmith
i got 12n = 24n - 3n..... they all have n's
jim_thompson5910
  • jim_thompson5910
\[\large \frac{ 4 }{ n } - \frac{ 1 }{ 2 } = -\frac{ 6 }{ 3n }\] \[\large \frac{ 2*4 }{ 2*n } - \frac{ n*1 }{ n*2 } = -\frac{ 6 }{ 3n }\] \[\large \frac{ 8 }{ 2n } - \frac{ n }{ 2n } = -\frac{ 6 }{ 3n }\] \[\large \frac{ 8 - n }{ 2n } = -\frac{ 6 }{ 3n }\] \[\large 3n(8 - n) = -6*2n\] \[\large 24n - 3n^2 = -12n\] does that help?
katherinesmith
  • katherinesmith
now what the heck am i supposed to do
jim_thompson5910
  • jim_thompson5910
\[\large 24n - 3n^2 = -12n\] \[\large 24n - 3n^2 + 12n = 0\] \[\large -3n^2 + 36n = 0\] \[\large -3n(n - 12) = 0\] how about now?
jim_thompson5910
  • jim_thompson5910
don't forget about checking your answers and keep the domain in mind
katherinesmith
  • katherinesmith
n = 0 and n = 12?
jim_thompson5910
  • jim_thompson5910
check both of those only one is the true answer
katherinesmith
  • katherinesmith
12
jim_thompson5910
  • jim_thompson5910
good
katherinesmith
  • katherinesmith
i have 2 left and they're super quick graph ones
katherinesmith
  • katherinesmith
What is the approximate value of the function at x = 0?
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jim_thompson5910
  • jim_thompson5910
you tell me
katherinesmith
  • katherinesmith
i don't know...4?
jim_thompson5910
  • jim_thompson5910
you're looking at the point when y = 0
jim_thompson5910
  • jim_thompson5910
how about x = 0
katherinesmith
  • katherinesmith
1.3 ?
jim_thompson5910
  • jim_thompson5910
close enough
katherinesmith
  • katherinesmith
last one! What is the approximate value of the function at x =5
1 Attachment
jim_thompson5910
  • jim_thompson5910
again tell me what you got
katherinesmith
  • katherinesmith
this one i really don't know.
jim_thompson5910
  • jim_thompson5910
find x = 5 on the x axis
jim_thompson5910
  • jim_thompson5910
draw a vertical line through x = 5 find where that vertical line crosses the red curve
katherinesmith
  • katherinesmith
-0.75?
jim_thompson5910
  • jim_thompson5910
then report the y coordinate
jim_thompson5910
  • jim_thompson5910
yes approximately
katherinesmith
  • katherinesmith
i got every single one right... i actually wanna cry. thank you beyond words. @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
you're welcome

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