f*g(x) and f/g(x) homework check

• Feb 12th 2010, 07:09 PM
snypeshow
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
• Feb 12th 2010, 09:20 PM
Soroban
Hello, snypeshow!

Looks good to me . . . Nice work!

• Feb 12th 2010, 09:38 PM
snypeshow
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)) ?
• Feb 13th 2010, 12:35 AM
Delaop
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.