Calculus Help

Originally posted by adam warlock
Agreed. But still...

True, it's not terribly difficult, but I think it's a nice one because it brings together multiple problem-solving strategies (taking the derivative, sketching the graph to determine the domain/range, polynomial factoring, checking your answer, etc...).

Now, why did I come up with an answer that didn't work, even though the value was in the domain of the function? I think the squaring had something to do with it, but I'm not entirely certain. Thoughts?

Originally posted by PBE6
True, it's not terribly difficult, but I think it's a nice one because it brings together multiple problem-solving strategies (taking the derivative, sketching the graph to determine the domain/range, polynomial factoring, checking your answer, etc...).

Now, why did I come up with an answer that didn't work, even though the value was in the domain of the fun ...[text shortened]... I think the squaring had something to do with it, but I'm not entirely certain. Thoughts?

Now, why did I come up with an answer that didn't work, even though the value was in the domain of the function? I think the squaring had something to do with it, but I'm not entirely certain. Thoughts?

Now we're looking for slopes equal to 3/2, so we let y' = 3/2:

3/2 = (1/2)x^(-1/2) + x

Multiplying by 2 to clear fractions, and grouping irrationals on one side:

2x - 3 = x^(-1/2)

Is the bit you got wrong:
3/2=(1/2)x^(-1/2)+x multiplying by 2
3=x^(-1/2)+2x which is equivalent to
2x-3=-x^(-1/2)

Edit: And yes the problem is more educational than I first thought. I didn't solve it all the way through in my head. I just made the derivative and equaled it to 3/2.

You could always just set it equal to 0 and graph it. Find the zero using the calculator's "zero" function.

Then again, you could just use your calculator's solver function. AP will accept answers accurate to three decimal places. I believe it is three.

Originally posted by PBE6
It's actually quite cleverly worded, I think. A good example problem, and slightly more difficult that the standard grinder.

Since the slope is just the first derivative of the function, we start there:

y = x^(1/2) + (1/2)x^2

y' = (1/2)x^(-1/2) + x

Now we're looking for slopes equal to 3/2, so we let y' = 3/2:

3/2 = (1/2)x^(-1/2) + x

Multiply ...[text shortened]... QRT(3))/2 works, but (2 + SQRT(3))/2 doesn't. Therefore, the answer is (2 - SQRT(3))/2.

I was really close. I was stuck after I got 4x^3 - 12x^2 + 9x - 1 = 0

It's not the math that's hard to me, it's just understanding the concept. Like when you should divide, or when you should take the derivative here and stuff like that.

Originally posted by Best101
I was really close. I was stuck after I got 4x^3 - 12x^2 + 9x - 1 = 0

It's not the math that's hard to me, it's just understanding the concept. Like when you should divide, or when you should take the derivative here and stuff like that.

Got any more? These are kinda fun. 🙂

It's not the math that's hard to me, it's just understanding the concept. Like when you should divide, or when you should take the derivative here and stuff like that.


You have to understand the meaning of the equations you are looking at. If you are dealing with slope, then you are dealing with the first derivative. If you have been given the derivative, then you just work with the equation given. If you are given the function itself, then you need to take the first derivative and deal with it.


The dividing part was just a technique of solving an equation given one of its zeroes. The dividing part was simply algebra, not calculus.

Calculus was setting up the equation:

3/2 = (1/2)x^(-1/2) + x

After that, it was all algebra.

Originally posted by PBE6
Got any more? These are kinda fun. 🙂

Of course 🙂.
Another one I had trouble with.
Find a and b if the tangent line to the function f(x) = x^4 + ax^2 + b at x=1 is y=2x + 3.

@Eladar: I think it may take me some time to get used to.

Work with the info.

You know the point in question because when x=1, y=2(1)+3. In other words, y=5. That means the point (1,5) is on both the line and the graph. More importantly, it is on the graph.

Now work with the slope. We know the slope at x=1 is 2 since the slope of the tangent line is 2. Remember, the slope has to do with the derivative, so we need f'(x)= 4x^3 + 2ax. Knowing the slope at x=1 is 2 means...

f'(1) = 2. Which means...

4(1)^3 + 2(a)(1) = 2

Now simplify:

4 + 2a =2

Solving we find that a = -1

Plugging -1 in for a in f(x) we now get...

f(x) = x^4 - x^2 + b


Since we know that the point (1,5) is on the graph we know that:

f(1) =5

Which means:

1^4 -(1^2) +b = 5

Simplified we get:

b =5


So a = -1 and b =5


I did this as I typed so I may have made a computational error, but the technique in sovling the problem is correct.

Originally posted by Eladar
What you are loooking at is actually a second derivative problem. f'(6) = F''(6)

When I took the second derivative and solved for f'(6) I got:

f'(6) = f''(6+h) - 2b-6ch

Perhaps there's a substitution for f''(h+6), but I'm not seeing it.

Wow dude, I hope you did not mess this kid up too much with that awful answer.

f'(6) = lim h->0 (f(6 + h) - f(6)) / h. So divide by h to get (f(6 + h) - f(6)) / h = a + bh + ch^2, and take h->0 to get f'(6) = a. It's the sort of calculus problem you get in like the first week of class.

Excuse me but exactly where did he define f(6+ h) - f(6)?

Originally posted by Eladar
Excuse me but exactly where did he define f(6+ h) - f(6)?

In his original post. He obviously made a typo, because this is exactly the type of problem that is given in the first week or so of calculus, to check whether or not the student understands what exactly a derivative is and how to take a limit.

According to Best101 he didn't know there was a difference. We cleared it up. If you'd bother to read the other posts, you'd know this.

Originally posted by Eladar
According to Best101 he didn't know there was a difference. We cleared it up. If you'd bother to read the other posts, you'd know this.

Right. He made a typo. I did read the whole thread.

And your other answer was spot-on. Although I've never before seen the notation you describe. But beyond basic calculus you don't see the prime notation a whole lot, either.

Function notation, what can I say. I don't think anyone really likes it. My kids would much rather use 'y'.

As far as the typo goes, I think it helped to clarify the point that when dealing with math, especially AP Calculus, you really need to be careful when it comes to notation.

Originally posted by Eladar
Function notation, what can I say. I don't think anyone really likes it. My kids would much rather use 'y'.

As far as the typo goes, I think it helped to clarify the point that when dealing with math, especially AP Calculus, you really need to be careful when it comes to notation.

I've just never seen anyone use F = f'.

Personally I prefer Leibiniz notation, although I'm starting to shift over to the use of commas to indicate partial derivative as is common with tensors.