Prove that if the real-valued function f is strictly increasing, then f is oneto-one.

See proof below

Step-by-step explanation:

Let x,y be arbitrary real numbers. We want to prove that if x≠y then f(x)≠f(y) (this is the definition of 1-1).

If x≠y, we can assume, without loss of generality that x<y using the trichotomy law of real numbers (without loss of generality means that the argument in this proof is the same if we assume y<x).

Because f is strictly increasing, x<y implies that f(x)<f(y). Therefore f(x)≠f(y) because of the trichotomy law, and hence f is one-to-one.

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

45

step-by-step explanation: