Potential exam questions: i.) Define: f : X → Y is 1:1 iff

ii.) Define: f : X → Y is NOT 1:1 iff

iii.) Prove that f : R → R, f(x) = is NOT 1:1.

i.) State the Intermediate Value Theorem:

ii.) See HW problems Use the limit definition of the derivative to find the derivative of f(x) =

Prove: If c is a constant and f is differentiable at x, then (cf)′(x) = c(f′(x)) I will choose two of the above for the following exam 1 question: #.) Choose one of the following (clearly indicate your choice). #A.) #B.)

Suppose 2x^{2}y − 3y^{2} = 4. Find y′′ 1.) First find y′: Easiest method is to use implicit differentiation. Take derivative (with respect to x) of both sides.

\frac{d}{dx}(2x^{2}y − 3y^{2}) = \frac{d}{dx}(4)

4xy + 2x^{2}y′ − 6yy′ = 0

Solve for y′ (note this step is easy as one can factor y′ from some terms):

y′(2x^{2} − 6y) = −4xy
y′ = \frac{−4xy}{2x^{2}−6y} = \frac{−2(2xy)}{−2(3y−x^{2})} = \frac{2xy}{3y−x^{2}}

Hence y′ = \frac{2xy}{3y−x^{2}} 2.) To find y′′, take derivative of y′: y′′ = (\frac{2xy}{3y−x^{2}} )′ = \frac{(2xy′+2y)(3y−x^{2})−2xy(3yy′−2x)}{(3y−x^{2})^{2}} Since y′ = \frac{2xy}{3y−x^{2}} ,

y′′ = ^{\frac{(2x(\frac{2xy}{3y−x^{2}} )+2y)(3y−x^{2})−2xy(3y(\frac{2xy}{3y−x^{2}} )−2x)}{(3y−x^{2})^{2}}}