If f(x)=-5+1 and g(x)=x^3, what is (g。f)(0)?

Question

whpalmer4 · Answer

Are you sure you have your f(x) correct?  I don't see an $x$ in it.

anonymous · Answer

Oh sorry yea it's f(x)=-5x+1

anonymous · Answer

@whpalmer4

whpalmer4 · Answer

If memory serves me correctly, that notation is equivalent to g(f(x)), right?

whpalmer4 · Answer

Yes, it is.  

So, we need to do 1 of 2 things:

Either find an algebraic expression for $(g\circ f)(x)$ and evaluate it at $x = 0$, or evaluate $u = f(0)$ and then evaluate $g(u)$

anonymous · Answer

Yes I believe so

whpalmer4 · Answer

Let's do both!

2) if $f(x) = -5x+1$, what does $f(0) =$?  $f(0) = -5(0)+1 = $

anonymous · Answer

1

whpalmer4 · Answer

Good.  Now what is $g(1) = (1)^3 = $?

anonymous · Answer

1

whpalmer4 · Answer

Good.  If we had a lot of these to do, it might be worth doing the algebra to find a direct expression for $(g\circ f(x))$.  Here's how we would do that.

$$f(x) = -5x+1$$$$g(x) = x^3$$Now we take the right hand side of our definition of $f(x)$ and rewrite our definition of $g(x)$, replacing $x$ wherever we encounter it on the right hand side of $g(x)$ with $f(x)$.

$$g(x) = x^3 = (-5x+1)^3$$
We could expand that to $$g(x) = -125x^3+75x^2-15x+1$$but it is probably just as convenient if not more so in the earlier form.

whpalmer4 · Answer

Let's check our work:
$$g(0) = (-5(0)+1)^3 = (0+1)^3 = 1^3 = 1\checkmark$$
$$g(0) = -125(0)^3 + 75(0)^2 - 15(0)+1 = 1\checkmark$$

whpalmer4 · Answer

Both of the new equations work at $x=0$, at least.  That's a necessary, if not sufficient, condition.  We could check another point both ways:

$$x = 1,~f(1) = -5(1)+1 = -4, ~g(-4) = (-4)^3 = -64$$
$$(g(f(1)) = (-5(1)+1)^3 = (-4)^3 = -64\checkmark$$$$g(f(1)) = -125(1)^3+75(1)^2-15(1)+1 = -125+75-15+1 = -64\checkmark$$So our equations work at 2 different points, I think it is safe to conclude our work is correct.

whpalmer4 · Answer

Any questions about how I combined equations? Or anything else?

anonymous · Answer

So is our answer 0?

whpalmer4 · Answer

No, our answer was 1.  

$$g(f(0)) = (-5(0)+1)^3 = (0+1)^3 = 1^3 = 1$$

Or $$f(0) = (-5(0)+1) = 0+1 = 1, ~g(f(0)) = g(1) = (1)^3 = 1$$