leozap1
  • leozap1
*Fan & medal* Please explain how I solve this. I have no idea how to do this, when I tried I ended up with -30x^2 - 6. Let f(x) = -5x + 3 and g(x) = 6x - 2. Find f * g and its domain. A.-30x^2 + 28x - 6; all real numbers except x = 1/3 B. -15x^2 - 8x - 12; all real numbers except x = 3/5 C. -30x^2 + 28x - 6; all real numbers D. -15x^2 - 8x - 12; all real numbers
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
probs a good idea to multiply the functions
leozap1
  • leozap1
That's what I did, and I got -30x^2 - 6
anonymous
  • anonymous
\[f*g=(-5x+3)(6x-2)\] \[f*g=-5x(6x-2)+3(6x-2)\] \[f*g=-30x^2+10x+18x-6\] \[f*g=-30x^2+28x-6\]

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anonymous
  • anonymous
now we need to find the domain.
anonymous
  • anonymous
we can use to methods to find the domain; graphical method or algebra
anonymous
  • anonymous
with me?
leozap1
  • leozap1
Yea I see what I did wrong thank you. I don't know how to find the domain though.
anonymous
  • anonymous
well here there are only two functions with our proposed solution right; answers A and C
anonymous
  • anonymous
A says all real numbers except x=1/3 and C says x has solutions for all real numbers
anonymous
  • anonymous
so we say to our selves, if we plug x=1/3 into the equation, would we get a value for the function?
anonymous
  • anonymous
Sure we do. f(x)=0 for x=1/3. but thats still a solution of the function right? that number exists. hence our answer is C
anonymous
  • anonymous
all g?
leozap1
  • leozap1
Thanks! This really helps me. I also have some questions similar to this, but with division. Could you help me with it? Here is one of the questions. Let f(x) = x^2 - 16 and g(x) = x + 4. Find f/g and its domain. (1 point) A. x + 4; all real numbers except x =/ 4 B. x + 4; all real numbers except x =/ -4 C. x - 4; all real numbers except x =/ 4 D. x - 4; all real numbers except x =/ -4
leozap1
  • leozap1
The =/ is suppose to be a does not equal to sign.
anonymous
  • anonymous
\[f/g=\frac{ x^2-16 }{ x+4 }\] \[f/g=\frac{ (x-4)(x+4) }{ x+4 }\] \[f/g=x-4\] and since the denominator of the previous step has x-4, then a function does not exist at a certain point when the denominator equals zero. therefore \[f/g=x-4\] \[\in R, \neq-4\]
anonymous
  • anonymous
hence answer is D
leozap1
  • leozap1
This one's a lot easier than the first, thank you.
leozap1
  • leozap1
Could you check my work on this last one? Let f(x) = -5x - 4 and g(x) = 6x - 7. Find f(x) + g(x). I got x-11 for this one.
anonymous
  • anonymous
looks good
leozap1
  • leozap1
Thanks!

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