Calc math homework help, please.

[MrDudeMan]

you are this far along in the semester and havent learned the derivative? im glad im not in your school system

 

[PowerMacG5]

Originally posted by: MrDudeMan
you are this far along in the semester and havent learned the derivative? im glad im not in your school system

Yeah, I agree with you. I am in AP Calc BC in my HS (full year course), and we learned what a derivitive was in the first week of school. Of course, we already knew the definition (lim (h-->0) [f(x+h)-f(x)]/h), but the teacher never called it what it actually was. Now, out of curiosity, I taught myself differential calculus over the summer, so I am cruising along so far. There is no way in a pre-calc course you did not learn the limit I just stated.

 

[Howard]

Ahh, never mind. I think i figured it out.

 

You can choose to learn derivatives in advance, but don't use that for this problem. Only use it to verify your answer. Many times, teachers don't appreciate you using a different method from what you're taught. There's a reason they want you to learn certain concepts. Trust me when I say derivatives will be all easy, but it's understanding limits that will be helpful to you as things get more complicated if you choose to go further in maths.

b = 8.

Here's how to work it using limits:

We're looking for this at the end: (Point slope equation) y - y1 = m (x - x1), since m is lim dx->0 (dy/dx)

f(x) = x^2 + 6x + 9. We want the equation to the line tangent to f(x) with slope of 4.

dx will stand for delta x and dy for delta y; m is the slope

The formula is m = lim dx -> 0 [f(x+dx) - f(x)]/dx

1) f(x+dx) = (x+dx)^2 + 6(x+dx) )+ 9 = x^2 + 2xdx + dx^2 + 6x + 6dx + 9

2) f(x+dx) - f(x) = (x^2 + 2xdx + dx^2 + 6x + 6dx + 9) - (x^2 + 6x + 9)
= x^2 - x^2 + 2xdx + dx^2 + 6x - 6x + 6dx + 9 - 9
= 2xdx + dx^2 + 6dx = dx (2x + dx + 6)

3) [f(x+dx) - f(x)]/dx = dx (2x + dx + 6)/ dx = 2x + dx + 6

4) lim dx->0 [f(x+dx) - f(x)]/dx = 2x + 0 + 6 = 2x + 6


So, m = 2x + 6

We are given a specific slope and asked to find the corresponding equation to the line tangent at that point. We want to find the points. At slope m = 4, we have:

4 = 2x + 6
=> 2x = -2 => x = -1

We were able to obtain the value of x this way. If x is -1, then we plug it into the original equation to get the value of y at x = -1:

y = x^2 + 6x + 9 = (-1)^2 + (6*-1) + 9 = 1 - 6 + 9 = 4. Hence y is 4 when x is -1. The corresponding point is (-1, 4).

We're back to solve the problem:

y - y1 = m (x - x1)

Hence, y - 4 = 4(x-(-1)) = 4(x+1) = 4x + 4 => y = 4x + 4 + 4
=> y = 4x + 8 (final answer!)

It's been a while since I took calculus. Hope this helps! I showed steps that could easily be skipped, just in case you wondered how I got from one place to another.