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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
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?
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
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
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
Great last one.If ƒ(x ) = 3x + 1, then ƒ(a + h ) - ƒ(a ) =
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.