# Thread: f*g(x) and f/g(x) homework check

1. ## f*g(x) and f/g(x) homework check

hey guys,
I just want to check if i did these questions right, ym textbook doesn't have any examples of them, so there's no way for me to check

a) f(x) = sqrt(x+1)
g(x) = sqrt(x-1)

f*g
= sqrt(x+1) * sqrt(x-1)
= sqrt[(x+1)(x-1)]
= sqrt(x^2 - 1)

f/g
= sqrt(x+1) / sqrt(x-1)
= sqrt[(x+1)(x-1)] / x-1

thanks

2. Hello, snypeshow!

Looks good to me . . . Nice work!

3. thanks!

I have another question that i'm having trouble with.

f(x) = 10^x
g(x) = log x

f(g(x)) = 10^logx
g(f(x)) = log (10^x)

how would i simplify f(g(x)) and g(f(x)) ?

4. First of all, I assume that $\displaystyle log = lg$ which is the common logarithm.

$\displaystyle f(x) = 10^x$
$\displaystyle g(x) = log(x)$

Now it is obvious that
$\displaystyle (f \circ g)(x) = f(g(x)) = f(log(x)) = 10^{log(x)} = x$ and
$\displaystyle (g \circ f)(x) = g(f(x)) = g(10^x) = log(10^x) = x$

This is followed by the definition of logarithms:
If $\displaystyle x = b^y$, then $\displaystyle y = log_b x$.

When we simply make the substitution $\displaystyle x = b^y$ in $\displaystyle y = log_b x$, we get $\displaystyle y = log_b b^y$.
We can also substitute $\displaystyle y = log_b x$ in If $\displaystyle x = b^y$ when we get $\displaystyle x = b^{log_b x}$.

We also note that since $\displaystyle (f \circ g)(x) = x$ and $\displaystyle (g \circ f)(x) = x$, f and g are each other's inverse functions.

By the way, do not forget about the range of definition when dealing with functions.