Calculus Help! Psychic powers required?

Jimmy King

I'm taking Calc 1 and we just started derivatives. This is homework, but it's ungraded and I've already got the correct answers, I just don't understand how those answers were determined. I don't see any explanation of a process for solving these in the book and the teacher didn't, so apparently I'm supposed to just already understand some relationship that I don't.

First I've got this: Find an equation of the line that is tangent to the graph of f and parallel to the given line.

f(x) = 1/(sqrt(x)) with a line of x+2y-6=0.
the derivative comes out to 1/(2x * sqrt(x)). That's fine, I worked that out myself.
The slope of that line is -1/2, I need the point at which that derivative = -1/2, I suppose, based on what the book shows for numbers but does not explain. That point is x=1.

The answer then goes on to say that means at (1,1) the tangent line is parallel to x+2y-6=0. How did we get here?
1 + 2y - 6 = 0
2y -5 = 0
2y = 5
y = 5/2
or
2y =-x + 6
y = -1/2x + 3
y = 5/2 if x =1.

Where did (1,1) come from? (1,1) along with the known slope of -1/2 is then put into point slope form to find the y=mx+b form for that line... that's fine, I get that, if I knew why (1,1) was picked

Then I've got this:

Identify a function f that has the given characteristics.
f(0) = 0; derivative of f(0) = 0; derivative of f(x) >= for all x != 0.
The book says "answers may vary" and gives f(x) = x^3. I can see how that fits, but why should I have known this? Should I "just know" or is there some process to finding answers for this that I haven't grasped?

musanman

Set the derivative equal to -.5 and solve to get x=1. The second one is a conceptual problem , I'd suggest sketching a graph and trying to think of functions that look like it

kime

The slope of that line is -1/2, I need the point at which that derivative = -1/2, I suppose, based on what the book shows for numbers but does not explain.

That's what derivatives are, graphically. When you take the derivative of that function, it will then tell you the slope of that function at any given point. (Technically, the slope of the tangent line on the function at any point.)

So I'm assuming you've got no problems solving for x=1, which is just algebra. Your question, then, is where did the point (1,1) come from? Well, remember, we are looking for a line tangent to that original function, which means it will be touching the function at that one specific point where the slopes match up. You found that the slopes match up at x=1, so you know your line will be touching (ie equal) to the original function at x=1.

So, what is the original function at x=1?

f(1) = 1/(sqrt(1)) = 1

So y=1, and the point is (1,1).

Your mistake was plugging x=1 into the line, which is not what we want. The line can be anywhere, it's position is completely and utterly unimportant. Only it's slope is important. For more explanation on that, see the spoiler right below.

So say you have y=3x. Obviously, the slope is 3, so any other line with a slope of 3 is parallel to that.

If y=3x+6, then that has the same slope, but a completely different position. The points on any of the two lines will never be the same.

So you are finding the tangent to the original function, but parallel to the line. You never, then, need to find any point on that line (ever), only the slope.

You had the right idea, but missed a bit of conceptualization, I think.


For the second problem, there are hints that you should pick up. f(0) = 0, that means there will be no constant term. So no f(x) = 3x + 1 or f(x) = x^2 - 5, but only things like f(x) = 3x or f(x) = x^2.

The derivative of f(0) = 0 takes out a lot of things you probably wouldn't consider, but one type of function you may have. What is the derivative of, say f(x) = 3x? Or f(x) = x? You should see that it won't be 0 at f(0), because when you take the derivative you get rid of the x. That means the power of x has to be bigger than 1.

Your last hint is incomplete . I think it was derivative of f(x) >= 0 for all x != 0? Based off the answer. Well, as I said above, the derivative tells you the slope of a function. So if it's always positive (except at x=0), what does that mean? The slope is always positive, so the function is always going up, ie always increasing (from left to right, of course, which is the "natural" way of looking at it usually). That is a big hint, and so you should be able to sketch out what some possibilities would be. What you should know are the general shapes of f(x)=x, x^2, x^3, x^4, etc. Look those up, and you'll see that whether the power is odd or even determines what the general shape looks like. So all odd powers look like they'll work, including f(x)=x^3, which is what the book gave. Make sense?


Let me know if you have any questions or if I didn't explain something well enough.

Jimmy King

Ahh, ok. I've got it for the first one now. I just didn't even really get what the question was asking because the teacher didn't show us any examples worded this way and with this given information and I couldn't find any examples with this wording in the section of the book that these questions are for practice for, either. It didn't click for me that setting the derivative to the already known slope and solving would give me x for the point because we just haven't looked at them like that. Everything we've done was plug the function into the difference formula and find the limit for delta X -> 0.

I guess the second one is "know your common functions and their graphs", really. Which I'm working on. My pre-calc teacher didn't make us memorize any of them (or a number of other things this teacher expects to have memorized... or show us certain others at all) which has put me at a big disadvantage in this class.

kime

Yeah, you should just try to practice some.

y = m * (x - h) ^ p + k (Sorry, I forget what the standard letters are for that). Know what happens when you vary k, know what happens when you vary h, know what happens when you vary p, know what happens when you vary m. Notice the minus sign on h, what does that mean? What happens if you throw minus signs in front of the other variables?

That's all stuff you should be familiar with, and that's just with some basic functions. There's also the trig functions, exponentials, etc. It's not hard, really, but since your previous math teachers didn't do a good job, it's just time that you should put in.

Jimmy King

Fortunately we did cover transformations, so I know that -h shifts right, +h shifts left, +k shifts up, -k shifts down, - slope flips over the x axis, etc. But we never had to memorize graphs of x^2 (although you see it enough that you end up knowing it anyway), x^3, sqrt(x), cubed-root(x), |x|, etc. I can knock those out quickly enough now, but I hadn't ever thought of them in terms of derivatives and what the slope is doing.

Never even saw the graphs of the trig functions, though. That's really annoying since this teacher will toss up a graph of cos, tan, etc. and just assume you know what's going on and why.

Well, back to homework and then some time on khan academy this weekend. Thanks for the help on these, hopefully I'll get to do enough of them that it sticks with me. I'm sure you'll be hearing from me again with more concepts the book and teacher assume I've put together without seeing it spelled out for me at least once.