3/5 and4/7 put these over a common denominator
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It helps to first clarify that the notation (f - g)(x) simply means f(x) - g(x). Given that, let's look at our f(x) and our g(x) here, and use their definitions to find their difference.
When we're taking (f - g)(x), we simply substitute the expression 3x + 1 for f(x) and the expression x² - 6 for g(x) to obtain:
Or, ordering the polynomial from highest power to lowest and combining the constants:
Edit: By request, here's what would happen if you had something instead like:
In this case, you'd have to *multiply* the two function expressions together. Here's what that would look like:
Using the distributive property, we can distribute the expression
to the terms 
and 
:
Distributing again, we get:
And we're done.
When we're taking (f - g)(x), we simply substitute the expression 3x + 1 for f(x) and the expression x² - 6 for g(x) to obtain:
Or, ordering the polynomial from highest power to lowest and combining the constants:
Edit: By request, here's what would happen if you had something instead like:
In this case, you'd have to *multiply* the two function expressions together. Here's what that would look like:
Using the distributive property, we can distribute the expression
Distributing again, we get:
And we're done.
Let p be: John goes to the beach
Let q be: He will go surfing.
Then in symbolic form, the argument becomes:
p ⇒ q
p
---------------------
∴ q
An argument is valid if the conjuction of the premises implies the conclusion.
p | q | p ⇒ q | (p ⇒ q) ∧ p | [(p ⇒ q) ∧ p] ⇒ q
---------------------------------------------------------------------\
F | F | T | F | T
F | T | T | F | T
T | F | F | F | T
T | T | T | T | T
The table above shows that the argument is a tautology.
Hence, the argument is valid