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anonymous
 one year ago
Suppose f(x) = 2x  1 and g(x) = x^2. Find (f+g)(x) and state domain.
anonymous
 one year ago
Suppose f(x) = 2x  1 and g(x) = x^2. Find (f+g)(x) and state domain.

This Question is Closed

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3(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
 one year ago
Best ResponseYou've already chosen the best response.0x^2 + 2x  1 I get that part, it's more so that I have trouble understanding how to find domain.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Rule: 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.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3There are no restrictions (eg: division by zero errors) to worry about

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay! Thank you. Can you help me w/ another problem?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Sure, go ahead

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find 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
 one year ago
Best ResponseYou've already chosen the best response.3I think you meant to say \[\Large h(k(x)) = \left(\log_{5}(x)\right)^2 + 3\log_{5}(x) + 1\] right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, I forgot the x's.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3as for the domain of h(k(x)), any ideas?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3you 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
 one year ago
Best ResponseYou've already chosen the best response.3so the domain is x > 0 which in interval notation is \(\Large (0,\infty)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, that makes sense. Thanks!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3how about k(h(x)) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would log5(x^2  3x + 1)'s domain be (inf, +inf)?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3hint: 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, (1, 1) and (2, 1).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, looking at the table tells me that.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3so because of that, log(x^2  3x + 1) = log(1) when x = 1 but log(1) leads to a nonreal number. So x = 1 is NOT part of the domain

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3there are other x values that aren't part of the domain for this reason

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be\[(\infty, 1] \cup [1, 2] \cup [2, \infty)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3what are the roots of x^2  3x + 1 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are there even any? Doing the big x, what is there that multiplies to 1 and adds to three?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3no, so you'll need to use the quadratic formula

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3x^2  3x + 1 does not factor the roots are decimal values

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3I graphed x^2  3x + 1 with a graphing calculator https://www.desmos.com/calculator/fp80fcskll click on the xintercepts and the coordinates will pop up. Tell me what values pop up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.00.382 2.618 I'm so mad b/c I got those before but thought they were wrong.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3notice 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
 one year ago
Best ResponseYou've already chosen the best response.3So the domain of log(x^2  3x + 1) is \(\Large (\infty, 0.382)\cup(2.618, \infty)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for all of your help. I think I understand what to do now.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3sure thing, glad to be of help
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