Suppose f(x) = 2x - 1 and g(x) = x^2. Find (f+g)(x) and state domain.

- anonymous

Suppose f(x) = 2x - 1 and g(x) = x^2. Find (f+g)(x) and state domain.

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- 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

x^2 + 2x - 1
I get that part, it's more so that I have trouble understanding how to find domain.

- 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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## More answers

- jim_thompson5910

There are no restrictions (eg: division by zero errors) to worry about

- anonymous

Okay! Thank you. Can you help me w/ another problem?

- jim_thompson5910

Sure, go ahead

- 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

I think you meant to say
\[\Large h(k(x)) = \left(\log_{5}(x)\right)^2 + 3\log_{5}(x) + 1\]
right?

- anonymous

Yeah, I forgot the x's.

- jim_thompson5910

as for the domain of h(k(x)), any ideas?

- anonymous

\[[1, \infty)\]

- jim_thompson5910

close

- 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

so the domain is x > 0 which in interval notation is \(\Large (0,\infty)\)

- anonymous

Okay, that makes sense. Thanks!

- jim_thompson5910

how about k(h(x)) ?

- anonymous

Would log5(x^2 - 3x + 1)'s domain be (-inf, +inf)?

- jim_thompson5910

no

- 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

Yes, (1, -1) and (2, -1).

- anonymous

Well, looking at the table tells me that.

- 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

there are other x values that aren't part of the domain for this reason

- anonymous

Would it be\[(-\infty, 1] \cup [1, 2] \cup [2, \infty)\]

- jim_thompson5910

what are the roots of x^2 - 3x + 1 ?

- anonymous

Are there even any? Doing the big x, what is there that multiplies to 1 and adds to three?

- jim_thompson5910

no, so you'll need to use the quadratic formula

- jim_thompson5910

x^2 - 3x + 1 does not factor
the roots are decimal values

- anonymous

5/2 and 1/2?

- 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

0.382
2.618
I'm so mad b/c I got those before but thought they were wrong.

- 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

So the domain of log(x^2 - 3x + 1) is \(\Large (-\infty, 0.382)\cup(2.618, \infty)\)

- anonymous

Thank you for all of your help. I think I understand what to do now.

- jim_thompson5910

sure thing, glad to be of help

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