Question

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

Answer 1

(f+g)(x) is the same as f(x) + g(x)

so you just add up the functions and combine like terms if possible

Answer 2

x^2 + 2x - 1

I get that part, it's more so that I have trouble understanding how to find domain.

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

Sure, go ahead

Answer 7

Find h(k(x)) and k(h(x)), and state the domain. h(x) = x^2 - 3x + 1 and k(x) = log5(x).

So, \[(\log_{5} )^{2} - 3(\log_{5} ) + 1\]

Answer 8

I think you meant to say

\[\Large h(k(x)) = \left(\log_{5}(x)\right)^2 + 3\log_{5}(x) + 1\]

right?

Answer 9

Yeah, I forgot the x's.

Answer 10

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

Answer 11

\[[1, \infty)\]

Answer 12

close

Answer 13

you can plug in numbers for x smaller than 1

like x = 0.5

but you cannot plug in x = 0 or anything negative

Answer 14

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

Answer 15

Okay, that makes sense. Thanks!

Answer 16

how about k(h(x)) ?

Answer 17

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

Answer 18

no

Answer 19

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?

Answer 20

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

Answer 21

Well, looking at the table tells me that.

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

x^2 - 3x + 1 does not factor

the roots are decimal values

Answer 29

5/2 and 1/2?

Answer 30

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

Answer 31

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

Answer 32

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 \(\Large (-\infty, 0.382)\cup(2.618, \infty)\)

Answer 33

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

Answer 34

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

Answer 35

sure thing, glad to be of help