# Simple Calculus

• Nov 16th 2009, 12:38 PM
Crell
Simple Calculus
My math class has just started getting into calculus and I have a few questions that I'd like to ask. Our school uses a program called TLM that spits out some random math problems from a huge database, and here are a couple of questions that I'm having trouble with.

Given the function $f(x) = 4x^2-1x+4$
evaluate $f(2+\Delta x)$

For this question should I just pop in $2+\Delta x$ for x?

EDIT:

$y=5x^2+6x+21$

The question then asks me what is $\Delta Y$ and $\Delta y / \Delta x$ when x changes from 1.6 to 2.2
For delta y I just subbed in the numbers and got 15

For finding the dy/dx I'm drawing a complete blank. I feel like it should be easy but I can't figure out what to do.
• Nov 16th 2009, 05:10 PM
Scott H
You are correct for the first question:

$f(2+\Delta x)=4(2+\Delta x)^2-(2+\Delta x)+4.$

Your answer to the second question also looks correct. To find $\frac{\Delta y}{\Delta x}$, we just divide the answer by $\Delta x=2.2-1.6$.

To find $\frac{dy}{dx}$, we evaluate what is called a limit:

$\frac{dy}{dx}=\lim_{\small \Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}.$

We can do this by simplifying the fraction and noting that certain terms containing $\Delta x$ approach $0$ as $\Delta x$ itself approaches zero, leaving us with the derivative of the function $f(x)$.

Hope this helps!
• Nov 17th 2009, 03:59 PM
Crell
Thanks for the answers Scott, but I'm still having a bit of trouble grasping the concept for the second part of my second question.

$y=5x^2+6x+21$ when x changes from 1.6 to 2.2.

Find $\frac {\Delta y} {\Delta x}$

I don't really understand what "rule" or equation to use to find the answer. It ended up being 25.

Again, any help is much appreciated!

EDIT: I found out this is the delta process, but I'm still attempting to wrap my head around it.