anonymous
  • anonymous
Suppose f(x) = 2x - 1 and g(x) = x^2. Find (f+g)(x) and state domain.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
(f+g)(x) is the same as f(x) + g(x) so you just add up the functions and combine like terms if possible
anonymous
  • anonymous
x^2 + 2x - 1 I get that part, it's more so that I have trouble understanding how to find domain.
jim_thompson5910
  • jim_thompson5910
Rule: the domain of ANY polynomial is the set of all real numbers. You can plug in any real number in for x, for any polynomial, and you'll get some real number out.

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jim_thompson5910
  • jim_thompson5910
There are no restrictions (eg: division by zero errors) to worry about
anonymous
  • anonymous
Okay! Thank you. Can you help me w/ another problem?
jim_thompson5910
  • jim_thompson5910
Sure, go ahead
anonymous
  • anonymous
Find h(k(x)) and k(h(x)), and state the domain. h(x) = x^2 - 3x + 1 and k(x) = log5(x). So, \[(\log_{5} )^{2} - 3(\log_{5} ) + 1\]
jim_thompson5910
  • jim_thompson5910
I think you meant to say \[\Large h(k(x)) = \left(\log_{5}(x)\right)^2 + 3\log_{5}(x) + 1\] right?
anonymous
  • anonymous
Yeah, I forgot the x's.
jim_thompson5910
  • jim_thompson5910
as for the domain of h(k(x)), any ideas?
anonymous
  • anonymous
\[[1, \infty)\]
jim_thompson5910
  • jim_thompson5910
close
jim_thompson5910
  • jim_thompson5910
you can plug in numbers for x smaller than 1 like x = 0.5 but you cannot plug in x = 0 or anything negative
jim_thompson5910
  • jim_thompson5910
so the domain is x > 0 which in interval notation is \(\Large (0,\infty)\)
anonymous
  • anonymous
Okay, that makes sense. Thanks!
jim_thompson5910
  • jim_thompson5910
how about k(h(x)) ?
anonymous
  • anonymous
Would log5(x^2 - 3x + 1)'s domain be (-inf, +inf)?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
hint: look at the graph of y = x^2 - 3x + 1. Are there points on that parabola that have y coordinates that are either 0 or a negative number?
anonymous
  • anonymous
Yes, (1, -1) and (2, -1).
anonymous
  • anonymous
Well, looking at the table tells me that.
jim_thompson5910
  • jim_thompson5910
so because of that, log(x^2 - 3x + 1) = log(-1) when x = 1 but log(-1) leads to a non-real number. So x = 1 is NOT part of the domain
jim_thompson5910
  • jim_thompson5910
there are other x values that aren't part of the domain for this reason
anonymous
  • anonymous
Would it be\[(-\infty, 1] \cup [1, 2] \cup [2, \infty)\]
jim_thompson5910
  • jim_thompson5910
what are the roots of x^2 - 3x + 1 ?
anonymous
  • anonymous
Are there even any? Doing the big x, what is there that multiplies to 1 and adds to three?
jim_thompson5910
  • jim_thompson5910
no, so you'll need to use the quadratic formula
jim_thompson5910
  • jim_thompson5910
x^2 - 3x + 1 does not factor the roots are decimal values
anonymous
  • anonymous
5/2 and 1/2?
jim_thompson5910
  • jim_thompson5910
I graphed x^2 - 3x + 1 with a graphing calculator https://www.desmos.com/calculator/fp80fcskll click on the x-intercepts and the coordinates will pop up. Tell me what values pop up
anonymous
  • anonymous
0.382 2.618 I'm so mad b/c I got those before but thought they were wrong.
jim_thompson5910
  • jim_thompson5910
notice how the portion between 0.382 and 2.618 is below the x axis the points on the parabola in this interval have negative y coordinates or the y coordinates are 0. So we have to kick this interval out of the domain Start with \(\Large (-\infty, \infty)\) and kick out that interval to end up with \(\Large (-\infty, 0.382)\cup(2.618, \infty)\)
jim_thompson5910
  • jim_thompson5910
So the domain of log(x^2 - 3x + 1) is \(\Large (-\infty, 0.382)\cup(2.618, \infty)\)
anonymous
  • anonymous
Thank you for all of your help. I think I understand what to do now.
jim_thompson5910
  • jim_thompson5910
sure thing, glad to be of help

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