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.