Recent content by karisrou

I still don't really understand. Do you think you can show it?

17. For what value of x is the graph of y = 2 / (4 - x) concave downwards? I found the first derivative = 2/(4-x)^2 And then the second 4/(4-x)^3 But I think I might have messed up somewhere in there

I'm kind of confused on step 2. How do you find the tangent?

6. The number c satisfying the Mean Value Theorem for f(x) = sinx on the interval [1,1.5]: So if the MVT is f(b) - f(a) / b-a .997 - .841 / 1.5 - 1 so .156 / .5 so .312 But that isn't the correct answer. Any thoughts?

f(b)-f(a) / b-a so (e-1) - (1-1) / (1 - 0) Which gives 1.718...

1. If c is the value defined by the mean value theorem, then for f(x) = e^x - x^2 on [0,1], c= I found the two end points as [0,1] and [1,e-1], so the average slope is .71828... is that the answer then?

2. Find the shortest distance from the point (1,4) to a point on the parabola y^2 = 2x Not really sure what to do here next. I'd imagine i might have to fuse implicit differentiation? But not really sure.

Yeah, everything looks right. Thanks very much for your help!

I've got D = 2rtx^2 - 6x + 9 + 64/x^2 I'm not sure as to how this minimizes?

so 2b - 128/b2 and then would you multiply everything by b2 to clear the denominator? giving you 2b3 - 128 = 0 So b = 3rt64 So 4? Then if b = 4 h must equal 2, per substitution into the original equation. So then the areas of the sides are 8, and the base is 16, so the total area...

Right, so b2 + 4b(32/b2) Which simplifies to b2 + 128b/b2 Which simplifies to b2 + 128/b

But aren't I supposed to substitute h into b2 + 4bh? and then solve for the critical point? Or would I just take the derivative of that? And then if that's what I'm supposed to do, do i take the derivative in relation to b or h?

Alright cool. I've got b^2 + 128/b^2 = min Should I take the derivative now?

1) Find the point on the hyperbola xy=8 closest to (3,0). I honestly, have no idea what to do. I seriously do not remember discussing anything like this in class, nor having any previous problems in homework. If anyone can give me a start or walkthrough, that would be fantastic!

1) A cardboard box of 32in^3 volume with a square base and open top is to be constructed. Find the minimum area of cardboard needed. Since it has a square base, length and width must be the same, variable b for base. So volume could be written as: hb^2=32 and surface area as b^2+4bh=min...