User:PseudonympH

Just to clarify, it's actually Pseudonym-pH. I tacked the pH on when regular "Pseudonym" was taken on Battle.net because it's, well, a science term. I also thought it was cool that it flowed well like that, so I kept the suffix. And no, as some people might think, I am not a female. :P

Math Class
If you're here because of the link posted on Nicenet, here's where you can find the nifty formulas and stuff. I did it here instead of there because i can do pretty math formatting like this:
 * $$\frac{x}{y}$$

Midyear review, number 11.
(Because it was the only one left :) )

For the function
 * $$f(x)=2x^2-\ln x$$

a. Find the equation of the tangent line to the graph of f(x) at (1,2).

First, we start off by finding the derivative of f(x)
 * $$\frac{d}{dx}(2x^2-\ln x)=4x-\frac{1}{x}$$

Taking this when $$x=1$$ will give us m for our y=mx+b equation. We can therefore conclude that m=3, and we can them solve for b:
 * $$y=mx+b$$
 * $$2=(3)(1)+b$$
 * $$b=-1$$

The equation of the tangent line is then y=3x-1.

b. Prove that f(x) is equal to 3 somewhere in the interval [1,2].

We can prove this easily using the Intermediate Value Theorem. First evaluate when x=1:
 * $$f(1)=2(1)^2-\ln 1=2$$

Then evalute when x=2:
 * $$f(2)=2(2)^2-\ln 2=8-\ln 2$$

Here we hit a bit of a snag, as we can't (easily) evaluate ln(2). However, since 2 is less than e, ln(2) must be less than 1. Therefore,
 * $$f(2) > 7$$

By the Intermediate Value Theorem, we can conclude that there is a number k on [1,2] such that f(k)=3.