If f(x)=x^3-3x^2-2x+5 and g(x)=2, then g(f(x))=?

Question

A.) 2x^3-6x^2-2x+10 B.) 2x^2-6x+1 C.)-6 D.) -3 E.)2

And what if it were f(g(x))? Help, I'm confused!

Answer 1

This is kind of a trick question.

g(x) = 2 means that the answer is always 2, no matter what value you use for "x"  So g (f(x)) = 2  

In the reverse case  f(g(x)), since g(x) is ALWAYS 2, just calculate

f(2) = 2^3 - 3*(2^2) - 2*2 + 5

     = 8 - 12 - 4 + 5

     =  - 3

Answer 2

the answer for the question is option E) 2

cause.. g(x) = 2 is a constant value.. whatever value of x, the value remains 2.. another reason is g(x) doesn't have any x term in the equation..

If it was f(g(x)) , then..

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

=> 8 - 3(4) - 4 +5

Answer: -3

there u go!.. enjoy :)

Answer 3

E) 2

g(x) = 2 always, no matter the value, even if you write F(x), its value is always 2!

If it were f(g(x)), then it would be substituying all "x"s by 2:

f(g(x)) = 2^3-3x2^2-2 * 2+5 = 8-12-4+5 = -3

Answer 4

if it were f(g(x)) it would be f(2) = 8-12-4+5 = -3

Answer 5

E.) 2

--------------------------------------...

And what if it were f(g(x))?

f(x)=x^3-3x^2-2x+5

g(x)=2

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

= (2)^3 - 3(2)^2 - 2(2) + 5

= 8 - 12 - 4 + 5

g(f(x)) = -3

Answer 6

G(x) is always 2 no matter what x is... So the answer is E)

And if it were f(g(x)=f(2) and then you put 2 where you have x in the function...

f(g(x))=2^3-3*2^2-2*2+5=8-12-4+5=-3

I hope you understand this, and good luck!