Thanks - Ashley

]]>I am not sure which property to use to write this as a single definite integral. I took a shot in the dark and thought about this problem as simply adding the net area under the curve of three different functions but this didn't help much. I guessed at the answer being the sum from -2 to 5 of f(x)dx. Im almost positive this is wrong, a hint may help.

Thanks ]]>

2(X-1)(x+3)^(2/3)+(2/3)(x+3)^(-1/3)(x-2)^2 and I found the critical values to be -3,-2,1 but when I tried getting the second derivative I got something crazy that I don't know how to set to 0. I got 12(X+3)^(-4/3)(X-1)-6(x-1)2 all over 9(x+3)^(-5/3) ]]>

2e^x -2e^(-x)/4

I set it equal to zero, but neither parts of the fraction can ever equal zero, right?

If this is the case, how do I tell when it is increasing or decreasing?

]]>9) If f''(x) doesn't ever equal 0, what steps should we take to describe the concavity?

EDIT: I figured this out too. -_-

]]>y= 6x+sin(3x)—-> I understand how to find the first derative using the chain rule and get

y'=6+ 3cos(3x)—> this is where I am forgetting how to use the rules to get the second derative the 6 is a constant so it goes away, but how do I deal with the 3(cos)3x I tried using the product rule then the chain rule but I am confusing myself.

Once I do get that derative how do I deal with that -sinx that I will eventually get

]]>**14)** The equation for the least material used would be $A=h(2\pi r)+8\pi r^2$, or (height)(diameter of the top / bottom) + (surface area of the top + bottom). Since the volume = 1, I made the secondary equation $1=h \pi r^2$ which can be solved for h: $h=\frac{1}{\pi r^2}$. Plugging that back into the original surface area equation, I got: $A=\frac{2\pi r}{\pi r^2}+8r^2$ or just $A=8r^2+\frac{2}{r}$.

Taking the derivative of that, I got $A'=16r-\frac{2}{r^2}$ or $\frac{16r^3-2}{r^2}$.

Set that equal to 0: $0=16r^3-2$ (the $r^2$ in the denom. is multiplied out).

Do some algebra and $r=\frac{1}{2}$.

I think the feasible domain would just be $[0,\infty]$, and the 1st derivative test shows that there's a minimum at $r=\frac{1}{2}$.

The area of the top or bottom of this cylinder would be $\pi\frac{1}{2}^2$ or $\frac{\pi}{4}$. The height would be $\frac{1}{\pi\frac{1}{2}^2}$ or $\frac{4}{\pi}$.

The ratio of height to radius would be $\frac{\frac{4}{\pi}}{\frac{1}{2}}$ or $\frac{8}{\pi}$ (which is the only part that I got right, implying that I did the calculations right for the wrong formula? or something).

**19)** With x being the length of the side of the square you're cutting out, the equation for the volume of the box would be $V=x(1-2x)(\frac{1}{2}-2x)$ or $\frac{8x^3-6x^2+x}{2}$.

The derivative of this would be $\frac{(24x^2-12x+1)(2)}{4}$.

Equal to 0: $0=24x^2-12x+1$.

$x=.3943, x=.1057$

Feasible domain: $[0,\frac{1}{4}]$, get rid of the first solution for x.

1st der. test, x is a local max.

$h=.1057$

$w=.2886$

$l=.7886$

I set up the equation for the area of the rectangle. A=xy. I also have that x is equal to 20, because the problem states that one of the sides of the rectangle is along the diameter. I'm not sure where to go from here. I know I need another equation but I don't know if I need the circumfrence of the semi circle or the area. I also don't know what to do with this second equation.

]]>I drew the cube and I know that on the base l=w which I set equal to x, then I know that the volume equals l*w*h or in this case V=x^2 h and since v=100 you can solve for h which is 100/x^2. So then I created another equation for the perimeter which i set equal to 4x+4(2x+2h) and then i substituted h from the first equation into the second equation. Then I solved for x but I didn't come very close to the actual answer so I don't really know where I went wrong. ]]>

I was doing 1. (C) and ran into the problem of one of my critical numbers being a negative root. What do I do with that? I thought you mentioned something about it in class, but I couldn't remember.

Also on 1. (D) How do I solve for the critical points of:

(1) ]]>$y=3x^2-(\frac{1}{x^2})$

$y'=\frac{6x^4+2}{x^3}$

So when x = 0, $y'$ wouldn't exist, right? And when $y'$ DNE at x, then x is a critical value?

Using 0 as a critical value, I'm getting that x < 0 as negative and x > 0 as positive, implying that there is a local min at x = 0, but the answer in the book doesn't support this.

EDIT: Looking over my notes again, I understand where my confusion is coming from, but I may still need some clarification. The picture you used (example a, a crazy looking function) has an asymptote where there is intuitively no max or min. But then soon after in the notes I have written down that x = c is a critical value when $f'(c)=0$ or DNE.

]]>Problem 14: It seems like more information is needed here but I really don't even know where to start. Staring at my picture doesn't help.

]]>And so that this post doesn't come off as just a useless vent, I humbly ask for help with problem #1. Are we supposed to be using the formula for the volume of a cylinder? So far I have that r = 20, dV/dt = -25, and dh/dt is what we need to find.

EDIT: Ugh, of course I figured it out right after posting this.

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