Just checking. If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find 2ƒ(1) + 3g(4).

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Just checking. If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find 2ƒ(1) + 3g(4).

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I came up with 42. Is that correct?
That's not what I got. What did you get for 2f(1) and 3g(4)?
5 and 37

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Other answers:

Hm. 2f(1) is 2*(f(1)) or 2 times the value of the function f(x) at 1.
So you should start by calculating f(1) and g(4), by plugging x=1 into f(x) and x=4 into g(x). Is that what you did?
Yes.
What was your final answer? Maybe I can plug it in and see if it is a match.
The final answer is 43. f(1) = 2 2f(1) =4 g(4) = 13 3g(4) = 39
I'm guessing x2 meant x^2 in your problem, not 2x.
right. did you do 13*3
Yes.
I think I got it thanks.
No problem. If you need to go through a few more I'll be happy to help.
Alright. If ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1, find [ƒ(2) - g(1)]^2.
Alright, start by calculating f(2) and g(1)
5 and 4
|dw:1441895933081:dw|
right
when does the exponent outside of the bracket come into play
5-4=1 1^2 is 1
Correct. The exponent "comes into play" when you calculate what's the value of the base. (that is the expression inside the [ ] parentheses.)
How about If ƒ(x ) = x^2 + 1, find ƒ(a + 1).
(a+1)^2 + 1
That's essentially correct.
You can write it out as a^2 + 2a + 2, if you want.
Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1) Could you use distributive So 3(a+1)+1 becomes 3a+3+1 then 3a+4
Yes.
Great last one.If ƒ(x ) = 3x + 1, then ƒ(a + h ) - ƒ(a ) =
3h
After you get the result f(a+1) = 3(a+1) + 1 you can treat the result as any other expression.
Great last one.If ƒ(x ) = 3x + 1, then ƒ(a + h ) - ƒ(a ) = 3h correct
Thank you. You have been such a great help.
Cool :)

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