If f(x)=x^2 + 3x + 1 and g(x) = x-1, then f (g(x)) is?

Question

please explain how to do it! i really dont get it!

Answer 1

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Since g(x) = x - 1, plug in x - 1 into the first equation to get f(g(x)).

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

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

f(g(x)) = x^2 + x - 1

Answer 2

All you have to do to get the answer for f(g(x)) is substitute g(x) where ever there is x in the original equation,
that is simply replace x in the equation with (x-1) to get the answer,

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

thus, f(g(x)) = x^2 + x - 1 ......... ANS

hope this helps!! :-)

Answer 3

x is the unknown value for the function f. (The letter f itself really doesn't mean anything [a function could be called h(x), g(x), etc. and the function would remain the same]).

So, for example, f(2) = 2^2 + 3(2) + 1 = 11.

We basically replace each "x" in f(x) with a 2.

Similarly, f(g(x)) means for each "x" in f(x), we replace it with "x-1".

So instead of x^2 + 3x + 1, we have (x-1)^2 + 3(x-1) + 1.

Expand and simplify from there.

Answer 4

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

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

Answer 5

f(x) = x² + 3x + 1
g(x) = x - 1

f ( g (x) )
= f (x - 1)
= (x - 1)² + 3(x - 1) + 1
= x² - 2x + 1 + 3x - 3 + 1
= x² + x - 1

Answer 6

Instead of x, substitute g(x) in the f(x) function equation.

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