f(x) = 15x - 12
g(x) = -15x2 + 14x - 10
find g(g(7))
my brain hurts - help me through the steps - functions for dummies version, I have 7 problems to do
THANK YOU
g(7) = -15*7^2 + 14*7 - 10 = -647
g(g(7)) = g(-647)
= -15(-647)^2 + 14(-647) - 10 = -6,288,203
Not hard - just be careful with the arithmetic.
I don't get why f(x) = 15x - 12 is there then.?
can you help with these - still having problems
15. f(x) = -13x2 + 12x + 14 g(x) = 11x - 13
find f(g(7))
16. f(x) = 11x + 15
g(x) = -11x3 - 15x2 + 13x - 13
find g(f(6))
17. f(x) = 15x3 + 13x2 - 15x + 15
g(x) = 15x + 11
find g(f(9))
18. f(x) = -13x2 - 13x + 14
g(x) = -13x - 11
find g(g(3))
19. f(x) = 15x + 12
g(x) = -10x + 15
find f(g(8))
is #15 -52,466???
15. yes
16.
f(6) = 11*6+15 = 81
g(81) = -11*81^3 - 15*81^2 + 12*81 - 13 = -5,943,307
17.
f(9) = 15*9^3 + 13*9^2 - 15*9 + 15 = 11,868
g(11868) = 15*11868 + 11 = 178,031
see? nothing to it!
Although all the big numbers are annoying. The arithmetic just gets in the way of the techniques.
Of course! Let's break down the steps to find g(g(7)) step by step:
Step 1: Substitute the input value (7) into the inner function g(x).
To find g(g(7)), we first need to calculate g(7). Using the function g(x) = -15x² + 14x - 10, we substitute 7 into the equation:
g(7) = -15(7)² + 14(7) - 10
Step 2: Simplify the inner function g(7).
g(7) = -15(49) + 14(7) - 10
g(7) = -735 + 98 - 10
g(7) = -647
Step 3: Substitute the result from Step 2 (-647) into the outer function g(x) again.
Now that we have g(7) = -647, we substitute this value into the function g(x) once more:
g(g(7)) = g(-647)
Step 4: Calculate the outer function g(-647).
Using the function g(x) = -15x² + 14x - 10, we substitute -647 into the equation:
g(-647) = -15(-647)² + 14(-647) - 10
Step 5: Simplify the outer function g(-647).
g(-647) = -15(419009) + 14(-647) - 10
g(-647) = -6,285,135 + (-9,058) - 10
g(-647) = -6,294,203
So, g(g(7)) = g(-647) = -6,294,203.
Remember to substitute the input values into the functions and simplify the expressions step by step. Let me know if you have any more questions or if there's anything else I can assist you with!