1. ## More calculus questions

here's some more problem that I am on right this moment:

1. determine the derivative of F(x)=(cos(2x-4))^3 at x=pie/b
2. determine d/dx 3x^4-3x/3x^4+3x
3. compute -3x^2x^3+3dx
4. an open rectangular box with volume 6m^3 has a square base. Express the surface area of the box as a function s(x) of the length x of a side of the base .
5. a box with an open top is to be constructed from a rectangular peice of cardboard with dmensions. b=5in. a=28in by cutting out equal squares of side x at each corner and then folding up the sides. find the maximum volume.
6. a spherical balloon with raduis r inches has volume 4/3pie^3. Find a function that represents the amount of air required to inflate the balloon from a radius of r innches to a radius of r+1 inches.
7. Lim as t approaches 2= t^2-4/t^3-8
8. absolute max. y=36-x (-6,6)
9. How many points of inflection in this graph
f(x)=12x^3=14x^2-7x-p

10. estimate the area from 0 to 5 under the graph
f(x)=81-x^2 using 5 rectangles from right end points.

thank you so much for the help!!!!! my life is on the line!!! i'm struggling in my AP calulus class and i have to pass this final. it's very urgent!!! thanks to anyone who helps.

2. 10. estimate the area from 0 to 5 under the graph
f(x)=81-x^2 using 5 rectangles from right end points.
First you determine the width of each rectangle by using this formula
$\Delta x = \frac{b-a}{n}$
Where n is the number of rectanges.
$\Delta x = \frac{5-0}{5} = 1$

$A = \Delta x(f(x1)+f(x2)+f(x3)+f(x4)+...)$

Since we are using right end-points triangles for $f(x) = -x^{2}+81$

$f(1)=80 , f(2)=77, f(3)=72, f(4)=65, f(5)=56
$

$A = 80 + 77 + 72 + 65 + 56 = 350$

3. Question#7

$\lim_{x \rightarrow 2}(x^{2} - \frac{4}{x^{3}} - 8) = 2^{2} - \frac{4}{2^{3}}-8 = -\frac{9}{2}$

Question#3

$\int -3x^{2} \sqrt{x^{3}+3}dx$

Let u = $x^{3}+3$

du = $3x^{2} dx$

-du = $-3x^{2} dx$

$\int -\sqrt{u}du = -\frac{2(u)}{3}^{\frac{3}{2}}$

Now you replace the u by $x^{3} +3$ and you get

$-\frac{2(x^{3}+3)}{3}^{\frac{3}{2}}$

4. thanks

5. For ur 4th question local maximum of the below equation.. I think we need to differentiate the equation and solve that for x by assigning the equation to zero

That means f ' (x) = 0

f(x)=x^3-9x^2+24x+4
3*x^2 - 18*x + 24 = 0
x^2 - 6*x + 8 = 0
(x - 2) * (x-4) = 0

either x = 2 or 4

so find out F(2) and f(4) and see which one is greater..

6. Originally Posted by calculus-geeks09
help
#8 - A function has an absolute maximum at c if f(c) is greater than or equal to f(x) for all x values in the domain which in your case is (-6,6)

#9 - get the second derivative of f(x)

d/dx f(x) = f'(x)
d/dx f'(x) = f"(x)

set f"(x) = 0 and solve for x, and that is your inflection point.

7. Originally Posted by fonso_gfx
#8 - A function has an absolute maximum at c if f(c) is greater than or equal to f(x) for all x values in the domain which in your case is (-6,6)
i don't get what you mean by this. can you explain it further pleasee.
i'm a dummie at this. LOL

thank u so much.