Calculus Help

Originally posted by Eladar
If the question is to find f(6) or F'(6), then you are correct. If the question is to find f'(6), then I'm correct.

Remember that a is a constant. If F'(6) equals a constant, then F''(6) = 0.

Originally posted by Eladar
It is only second derivatives of functions defined in this way that are zero, at least the second derivatives at specific numbers.


Best101,

As for the limit of x/(x-1) being 0/0, the common problem is that people do not reduce before replacing h with zero. Always reduce to cancel out the h in the denominator before substituting the 0.

Originally posted by adam warlock
F(x+h)-F(x)=(x+h)/(x+h-1)-x/(x-1)=[(x+h)(x-1)-x(x+h-1)]/[(x+h-1)(x-1)]=
=(x^2-x+hx-h-x^2-xh+x)/[(x+h-1)(x-1)]=-h/[(x+h-1)(x-1)]=

Now [F(x+h)-F(x)]/h=-1/[(x+h-1)(x-1)]

And taking the limit as h go to 0 we are left with F'(x)=-1/(x-1)^2

Originally posted by Best101
Thanks. This stuff makes Algebra look like nothing.

@joe shmo: Btw I'm just started Ap Calculus about a month and a half ago in High School.

Edit: Is there a difference b/w F(x) and f(x)? I didn't think it mattered.

Originally posted by Eladar
Yes, there is a very big difference between F(x) and f(x). f(x) is the derivative of F(x).

Originally posted by flexmore

Originally posted by Best101
I think I need to get out of Calculus, cause this stuff just doesn't make any sense.

Can anyone answer this one. Also do you think this is an appropiate test question for someone who just started Calculus about a month ago?

Q: Find the x coordinates of all the points on the curve x^1/2 + 1/2x^2, besides x=1, at which the slope of the tangent line equa ...[text shortened]... it if you don't want to because know how annoying it could be to type numbers on a computer.

Originally posted by adam warlock
Just think about what the slop of a curve at a given point as to do with its derivative and this exercise is piece of cake.


[b]Also do you think this is an appropiate test question for someone who just started Calculus about a month ago?
I think that you need to study a little bit harder[/b]

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.

Originally posted by PBE6
lightly more difficult that the standard grinder.

Originally posted by uzless
If my fist was travelling at mach 1.5 on an inverse tangental plane perpendicular to your face at a distance of 1 metre, what would be the expected kip force upon impact?