if f(x)=5xsquared-1 and g(x)=3x-1 find g(f(1)).?

Question

if f(x)=5xsquared-1 and g(x)=3x-1 find g(f(1)).
how do i solve this and whats the name of this kind of problem???

- thanksss

Answer 1

ok, it's a composite function

first solve for the inner portion .... f(x) = f(1)...so replace x with 1 in the equation f(x)
f(x)=5x^2-1...
f(1)=5(1^2)-1...=5-1...=4 so f(1)=4

now you can find g(x) because f(1) is equal to the x in g(x)...so if
f(1)=4 then replace x with 4 in the g(x) equation.

g(x)=3x-1....
g(4)=(3*4)-1...=12-1...=11

so g(f(1))=11


...It's pretty much a substitution. Since you know you are looking to use the equations for either f(x) or g(x) find which one you can do first according to g(f(1)). You can't do g(x) first because x=f(1) and you don't know what f(1) is yet. You can do the f(x) first because you can see you need f(1) so x=1 in the f(x) equation.

Answer 2

These are often called composite functions. One way to solve it is to first find f(1), and then use that to find your second answer. For example, here, f(1) = 4. So then take g(4).

Another way is to simply plug f(x) into g(x) where ever there is an x in g(x). Thus, you have

f(x) = 5x^2-1. So then g(f(x)) = g(5x^2-1) = 3(5x^2-1)-1.

Either way, you will arrive at the same answer.

Answer 3

It's called a 'composition of functions' and you solve it by substitution.

First, get f(1) = 5*1^2 -1 = 4
Then get g(4) = 3*4-1 = 11

And you're done.


Doug

Answer 4

f(x) == 5x^2-1, g(x) ==3x-1
g(f(x)) == 3f(x)-1 == 3(5x^2-1)-1 == 15x^2 - 4
g(f(1)) == 15*1^2 -4 == 15*1 - 4 == 15-4 == 11

The answer is 11.

This is called Composing Functions.

Answer 5

f(1) = 5 (1)^2 -1 = 5 -1 = 4
g(5) = 3(4) - 1 = 12-1 = 11
The answer is 11

Answer 6

f(x) = 5x^2 - 1
g(x) = 3x - 1

f(1) = 5(1)^2 - 1 = 5(1) - 1 = 5 - 1 = 4

g(f(1)) = g(4)

g(4) = 3(4) - 1 = 12 - 1 = 11

g(f(1)) = 11

Answer 7

you have to put the equation for F(x) into the G(x) equation... so G(x)= (3(5x^2)-1).. and you should be able to solve from there. i dont remember what this is called... er. something or other. hope i was of help!