Define: f : X → Y is 1:1 iff \underline{f(x) = f(y) implies x = y}

Define: f : X → Y is NOT 1:1 iff there exists x ≠ y such that f(x) = f(y)

Prove that f : R → R, f(x) = x^{4} is NOT 1:1.

Define: f : X → Y is onto iff \underline{f(X) = Y}.

Define: f : X → Y is NOT onto iff \underline{there exists y ∈ Y} s. t. there does not exist an x ∈ X s. t. f(x) = y (i.e., y is not in the image of f).

Prove that f : R → R, f(x) = x^{4} is NOT onto.

State the Intermediate Value Theorem: Suppose f continuous on [a, b], f(a) ≠ f(b) and N is between f(a) and f(b), then there exists c ∈ (a, b) such that f(c) = N.

Use the Intermediate Value Theorem to show that there exists c ∈ [1, 2] such that c^{4} = 2.

Use the Intermediate Value Theorem to show that there exists c ∈ [−2, −1] such that c^{4} = 2.

Defn: lim_{x→a}f(x) = L if for all ϵ > 0, there exists a δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ϵ Prove lim_{x→2}3x = 6 Take ϵ > 0. Choose δ = \frac{ϵ}{3} Suppose 0 < |x − 2| < \frac{ϵ}{3} Claim |3x − 6| < ϵ Proof of claim: |3x − 6| = 3|x − 2| < 3(\frac{ϵ}{3}) = ϵ Define f continuous at a:

Define f differentiable at a:

Prove that if f differentiable at a, then f is continuous at a. Hint: Show lim_{x→a}[f(x) − f(a)] = 0

Thm: If f is differentiable at x and c is a constant, then (cf)′(x) = c(f′(x))