SOLUTION: For this problem, let f(x) = 3x^2 - 1 and g(x) = 2x + 5.
Find f(g(-4)) and g(f(-4)).
The books answer is 26;99. I'm not sure how to solve this. Please show me how it's done.
Algebra.Com
Question 982998: For this problem, let f(x) = 3x^2 - 1 and g(x) = 2x + 5.
Find f(g(-4)) and g(f(-4)).
The books answer is 26;99. I'm not sure how to solve this. Please show me how it's done.
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
f(x) = 3x^2 - 1
f(x) = 3(x)^2 - 1
f(g(x)) = 3(g(x))^2 - 1 ... replace every x with g(x)
f(g(x)) = 3(2x + 5)^2 - 1 ... replace the g(x) on the right side with 2x+5
f(g(-4)) = 3(2(-4) + 5)^2 - 1 ... replace every x with -4. Now evaluate
f(g(-4)) = 3(-8 + 5)^2 - 1
f(g(-4)) = 3(-3)^2 - 1
f(g(-4)) = 3(9) - 1
f(g(-4)) = 27 - 1
f(g(-4)) = 26
I'll let you do the other part.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
(x)\ =\ f\left(g(x)\right)\ =\ 3\left(g(x)\right)^2\ -\ 1\ =\ 3\left(2x\ +\ 5\right)^2\ -\ 1\ =\ 3\left(4x^2\ +\ 20x\ +25\right)\ -\ 1\ =\ 12x^2\ +\ 60x\ +\ 75\ -\ 1)
To find \right))
, just substitute -4 in place of 
and do the arithmetic.
The other problem works the same way.
John
My calculator said it, I believe it, that settles it
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