OMG help me PLEASE, ap calc test teusday and i dont get this!

[Jittles]

EDIT: made title caps

I am so stressed this weekend. AP English test tomorrow, calc after that, essay due, physics test, ARGH! and I don't get this calculus.

I have about half this practice test done, but i'm totally stumped by these. Please help me figure this out. I know it isn't that difficult of calculus problems either.
THANK YOU!



This first one I just don't seem to get what I'm supposed to do. any ideas?
(d) Consider the family of functions defined by y = bxe^(bx) where b is a nonzero constant. Show that the absolute minimum value is the same for all nonzero values of b.

4. Let f be a function with f(1) = 4 and slope is given by (3x^2 + 1) / (2y)

(b) write an equation for the line tangent to the graph of f at x = 1 and use it to approximate f(1.2)


6. Consider the curve defined by 2y^3 + 6x^2y - 12x^2 + 6y = 1

(a) is show how to get dy/dx but I did that and got (4x - 2xy) / (x^2 + y^2 + 1) that is the correct answer but that might help part (b)

(b) write an equation of each horizontal tangent line to the curve.

(c) the line through the origin with slope -1 is tangent to the curve at point P. Find x,y of point P. I can do this once I know how to do part (b)

OK I'm out of breath...typing... whatever... You don't have to do them for me but point me in the right direction please!

Thank you so much for helping me!

 

[cchen]

<< I am so stressed this weekend. AP English test tomorrow, calc after that, essay due, physics test, ARGH! and I don't get this calculus.

I have about half this practice test done, but i'm totally stumped by these. Please help me figure this out. I know it isn't that difficult of calculus problems either.
THANK YOU!



This first one I just don't seem to get what I'm supposed to do. any ideas?
(d) Consider the family of functions defined by y = bxe^(bx) where b is a nonzero constant. Show that the absolute minimum value is the same for all nonzero values of b.

4. Let f be a function with f(1) = 4 and slope is given by (3x^2 + 1) / (2y)

(b) write an equation for the line tangent to the graph of f at x = 1 and use it to approximate f(1.2)


6. Consider the curve defined by 2y^3 + 6x^2y - 12x^2 + 6y = 1

(a) is show how to get dy/dx but I did that and got (4x - 2xy) / (x^2 + y^2 + 1) that is the correct answer but that might help part (b)

(b) write an equation of each horizontal tangent line to the curve.

(c) the line through the origin with slope -1 is tangent to the curve at point P. Find x,y of point P. I can do this once I know how to do part (b)

OK I'm out of breath...typing... whatever... You don't have to do them for me but point me in the right direction please!

Thank you so much for helping me! >>



d. since to find the min you take a derivative, the constant goes away

b. differentiate the function

b. set the top =0, bottom =0

 

[BillGates]

You need to start out with easier problems such as 2+8= ..... ah, screw it. We don't know.

 

[Jittles]

 

[Jittles]

help help help help help

please?

 

[zephyrprime]

d.) by y = bxe^(bx)
I'm not sure what you mean by by y. Does this mean b * y * y ?
This equation's minimum absolute value is 0. Since we're only considerin absolute values, the values minumum value must >= 0. Since the function equals 0 at x=0, this functions minimum abs value is 0. It's zero for any value of b because 0*anything = 0.

4.b You should know this one. The dirivative of any line function is it's slope. So (3x^2 + 1) / (2y) is the slope of of f() at x. With the slope of the funtcion know, you only need to determine the y offset of the function to get the final answer. The offset of the function is already given for f(1) but we need it for f(0).

remember the line equation:
y = ax + b

for this problem,
x = 1
y = 4
a = (3x^2 + 1) / (2y) with x=1, y=4

so now b can be solved for easily.
Now we have an equation for the tangent of f() at x=1. This equation is an approximation for f() for all values of x near x=1. The tangent of any line is an approximation of the line near the tangent point.

 

[zephyrprime]

6.b
The horizontal tangent lines are all of the form y = n where n is a real number. You see, all the lines are horizontal.

The points where there are horizontal tangents are the points on the line that are the peaks of "hills" and the very bottom of "valleys". Since these tangents are horizontal, they have a slope 0. So the horizontal tangents of f() occur wherever the slope of f() = 0. I hope you learned this is school because if you've only just learned this from me you're gonna be screwed in the AP test tomorrow.

Anyway, the derivative (4x - 2xy) / (x^2 + y^2 + 1) is the slope and it equals 0 whenever (4x - 2xy) = 0 because 0/anything = 0 except 0/0 which is undefined.

4x - 2xy = 0 when y = 2.

So substitute y=2 into 2y^3 + 6x^2y - 12x^2 + 6y = 1 and solve for x. If you do this, it seems that there is no value of x when y=2 so it seems that there aren't any tangents of this function that are horizontal.

 

[rgwalt]

For your first problem, take the derivative and set it to zero. Solve for x. Then you have to check the second derivative to show that your extrema is indeed a minimum.

Ryan

 

[BillGates]

I know all the answers, but I hate "OMG" and refuse to help.