math hw question? functions?

Question

evaluate g[f(2)] and f[g(-3)] for each pair of functions

f(x)=3x, g(x)=-x+1



i know the answer is -5 and 12, but please does anyone know howw to solve this? i dont get functions at all. thankss

Answer 1

for g[f(2)] , you plug the 2 into the f(x) function to get 3(2) which is 6.
You now plug that 6 into the g(x) function to get -(6)+1, which simplifies to -5

Hope that helps with the second one
Watch out for the double negative

Answer 2

A function f(x) simply expresses a value that something takes in relation to x.
f(2) simply means: "the value when x=2"

g(x) is another function where we use the letter g instead of f, to show it is a different function. But the concept stays the same.

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Easy way (two steps)

g(f(2))
1st step: f(2) = 3(2) = 6
2nd step: g(f(2)) means g(6) = -(6) +1 = -5

g(-3) = -(-3)+1 = +4
f(g(-3)) = f(4) = 3(4) = +12

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The "real" way.

(Doing the second problem first)
same concept. f(x) is the value relative to x.
What if x = g(x) ?

then f(g(x)) = 3*g(x)
(we simply used x = g(x) instead of a number)
which we can write out using the equation for g(x) = -x+1

We get f(g(x)) = 3*(-x+1)
If x = -3, then f(g(-3)) = 3*(-(-3)+1) = 3*(3+1) = 3(4)=12

the first problem.
g(x) = -x + 1
what if x = f(x) ?
g(f(x)) = - f(x) + 1 = - 3x + 1
for x=2
g(f(2)) = - 3(2) + 1 = -6 + 1 = -5

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Both ways (easy or real) are correct ways to solve this kind of problem. In almost all cases, both ways will work just as well.

There are some strange cases when combining the functions (or not combining them) leads to strange operations, like dividing by zero or taking the logarithm of a negative number. In such cases, you must take the way that avoids the impossible operation.

And there are cases where combining the functions is simply unsolvable.

Answer 3

Functions might look pretty ugly at times but evaluating them is as easy as substituting the unknown (the x in this example) with the values they are asking you to evaluate.

1) f(2) = 3 * 2 = 6 . Now subsititute that result into g[f(2)] = g(6) = -6+1 = -5

2) Same idea but different way around. First evaluate g(-3) = -(-3) + 1 = 3 + 1 = 4. Now substituting 4 into f[g(-3)] = f(4) = 3 * 4 = 12.

Tada!

Answer 4

functions are basically short hands for equations. y=3x can be expressed as f(x)=3x. you'd insert anything in the parenthesis (or brackets) into place for x.

you'd write g[f(x)] as -f(x)+1
f(x) is 3x, you'd write -3x+1.

Making sense?

So f(2) = 3(2), which is 6
g(6) is -6+1
g[f(x)]=-5

g(-3) is -(-3)+1, which is 4
f(4) = 3(4) which is 12.

Answer 5

g[f[2]] f[g[-3]]
first you plug in the x value
g[3(2)]=g[6]
then you take that and put it in the g function
-6+1=-5
then you repeat for the next problem
f[-(-3)+1]
f[4]
4(3)
12
if you need any more help let me know

Answer 6

For g(f(x)) = g(x) = -f(x)+1 = -3x +1 --> x = 2, g(f(2)) = -5

f(g(x)) = 3*g(x) = -3x+3, x = -3, f(g(-3)) = 12

So in f(g(x)) you replace x in f(x) with g(x) and simplify. Same with g(f(x))

Answer 7

ok, so in case you divide the 1st and 2d words, as an occasion, -.6/-.35. I used g(x) yet for f(x), it could be one million.38/one million.3 Now take the 2d and third and divide interior the comparable way. If the effect is the comparable, that's linear.