First-Year Calculus Problems... :(

[25thhour]

Sooo, I am having immense trouble with some (all) of my assignments due to the fact that I am having serious family troubles

I totally have no idea how to do these questions and 8 of these assignments makes up 20% of my final mark...

K so here it goes -

QUESTION 1:
Find all points on the graph of y=x^3+8x^2+5x where the tangent line is horizontal.

I think for this one I find y' and set it = to 0 then solve for the values of x?
Heres my work:
y' = 3x^2+16x+5
y' = (3x+1)(x+5) = 0
x = -1/3, -5

QUESTION 2:
This one effs me up because I hate trig identities...

Find the x-coordinate of all points on the curve:
y = 10xcos(5x)-25sqrt(2)x^2-97 , pi/5<x<2pi/5
Where the tangent line passes through the point (0,-97) (not on the curve)

For this one, I honestly was totally lost after I took the derivative (right I think???)

y' = 10cos(5x) + (10x(-sin(5x)+5cos(x)))-50sqrt(2)

would I put my original equation into the slope formula:
m = (y2 - y1)/(x2 - x1)
to give me:

[10xcos(5x) - 25sqrt(2)x^2 - 97) - 97] / x2 - 0

then set this formula equal to y' to find the m'

[10xcos(5x) - 25sqrt(2)x^2 - 97) - 97] / x - 0 =
10cos(5x) + (10x(-sin(5x)+5cos(x)))-50sqrt(2)

I then multiplied both sides by x to get rid of the fraction.

I don't know how to go any further... I am terrible with trigonometric functions.

QUESTION 3:

Let y = [F(tan(6x))]^6 for some differentiable function f. Determine dy/dx at x = 0 given that f(0) = 3 and f'(0) = 1/81

sooo I think I got this bitch in the bag... I hope...

So first off I did implicit differentiation to find:

[6f(tan(6x))]^5 (f'tan(6x) + f(6sec^2(6x)

I then replaced the f with 3 and the f' with 1/81 and x with 0 and I get an answer of 18. Right? or wrong?

QUESTION 4:

Determine Values of A and B so that the function below is differentiable at x = -4.

F(x) = ax^2 - 7x + 81 x< or equal to -4
bx+1 x > -4

So first I found a general equation by making the 2 equations = then, I guess taking the limits from -4 from the negative and -4 from the positive of the appropriate equations and setting them equal to each other to get

16a + 4b = -108

I then took the derivative of the first equation to find a:
I found a to be -7/8 and then plugged it into the general equation I found to get b = 23.5

Thanks guys! Im in desperate need of help i'll send a generous amount of credits to whoever helps me

[Tidus810]

I haven't had to do any of this recent enough to be 100% sure about how your work came out, but a lot of it looks correct. I can tell you, though, that for part of the derivative for Q2 you made a boo boo:
When you did the Chain Rule for 10xcos(5x), you got for the second term:
(10x(-sin(5x)+5cos(x)))
when actually I think it's:
-50xsin(5x)
which comes from 10x * -5sin(5x)

Hope this helps, if only a tiny bit.

[25thhour]

Could someone halp me work through these to see if im correct?

[jgano]

It's been a while since I took calculus; so far, I've done the first two problems.

With the first one, your x values are correct; now, you just have to find the corresponding y values.

With the second one, Tidus was right about the chain rule. Also, you're subtracting -97, NOT positive 97. You should have a slope problem of [(10*x*cos(5x) - 25*sqrt(2)*x^2 - 97) - (-97)]/[x - 0] = 10*cos(5x) - 50*x*sin(5x) - 50*sqrt(2)*x. The 97's will cancel out, and you can then divide the left side by x, and you're left with 10*cos(5x) - 25*sqrt(2)*x = 10*cos(5x) - 50*x*sin(5x) - 50*sqrt(2)*x. Then, you can cancel the 10*cos(5x) terms, add 25*sqrt(2)*x to both sides, and factor out the x's to get -50*sin(5x) - 25*sqrt(2) = 0. The rest should be finding out what x is equal to.

EDIT: Okay, for question 4, for F to be differentiable at x = -4, the 2 equations have to be equal at x = -4 and the slopes of the 2 equations have to be equal at x = -4. So, you have to take the derivatives of both equations of F and set them equal to each other. This will leave you with 2 equations using a and b.

I had an explanation for question 3, but I deleted it because I wasn't sure if I did it right myself. Sorry.

[reuben_tate]

I can probably help over skype or something if you need help? I know how looking at a whole bunch of text at once can be intimidating sometimes.

[25thhour]

Originally Posted by reuben_tate

I can probably help over skype or something if you need help? I know how looking at a whole bunch of text at once can be intimidating sometimes.

That'd be awesome!!! ibeatniggs is my skype :P Walker Adair is the name on it

[reuben_tate]

That was a fun skype session Objective complete...huzzah! ( ^)>!