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1)

If you graph these two graphs you see they intersect at x=1.

As k>1 then the vertical line x =k is to the right of the intersection.

A) To find the area you need to integrate from 0 to k.

(0 to 1) 3x dx + (1 to k)3/x dx

= x (0 to 1) + 3lnx (1 to k)

= 1 + 3ln k

B) If R is 7 then7 = 1 + 3ln k

6 = 3ln k

2 = ln k

k = e

C) dR/dt = 5 given find dk/dt at k=15

Using R from above: dR/dk = 3/kNow dk/dt = (dR/dt) / (dR/dk) = 5/(3/k) = 5k/3

So dk/dt (when k = 15) is 25 units per second

2) Hi, Joe,

t

if v(t) = --------

1 + t^2

Here is what you will do: [You didn't show any of the work you already did, so I assumeyou just don't know how to approach it.]

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(a) Max velocity. Differentiate v(t), using quotient rule. Set that v'(t) (just the top) equalto 0 and solve for t. Put that back into v(t) for your max.

(b) Position. INTEGRATE your v(t) to get x(t). You get a '+ C', of course. Put t=0, x=5into the equation for x to solve for C. THEN put t = 6.

(c) Find lim[t->infinity] t/(1+t^2). I think it will be zero.

(d) Set x(t) = 500 and see if you get a solution for t. If you do, say YES.

3)

a) Since we know that Sin(kx) = 0 when kx =

=> when x = /k your sine function hits it's zero.

So in part a), the area of the region bounded by one arch of the graph, is the integral from

0 to /k of kSin(kx) (from 0 to /k) kSin(kx) dx

- Cos(kx) ( evaluated from 0 to /k )

- Cos( ) + 1

2.

So your area is 2 for any value of k.

b) First we have to find the tangent line when your graph is at a zero.

f ' ( x ) = k Cos(kx)

Since your zero is obtained at /k, evaluate your derivative there.

f ' ( /k ) = k Cos( )

f ' ( / k ) = - k

this is the slope of your tangent line. and we know one coordinant of that line is ( /k , 0 )

Using those two pieces of information, we can find the equation of the line given by thefollowing formula

y - y0 = m( x - x0) where ( x0 , y0 ) = ( /k , 0 ) , and m = - k, plugging these in we get

y - 0 = - k ( x - /k )

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Solve the p'(q) = -3 for q and you have all three values of q

------------------------------------------------------------------------------------------------------------(a) P(x)= x^4+ ax^3+ bx^2+ cx + d----(1)

Since, the function P(x) has a relative maximum at (0,1).Therefore, the function P(x)passes through (0,1).

Then, 1=0^4+a(0)^3+b(0)^2+c(0)+d

d=1

Therefore (1) becomes

P(x)= x^4+ ax^3+ bx^2+ cx + 1---(2)

Differentiating equation (2) with respect to x, we get,

P^' (x)=4x^3+3ax^2+2bx+c

Since, the function P(x) has a relative maximum at (0,1).

Then, P^' (x)=0 at x=0

Then, (2) becomes,

0=4(0)^3+3a(0)^2+2b(0)+c

c=0

Then, (2) becomes,

P(x)= x^4+ ax^3+ bx^2+ 1---(3)

Given: P(x) is symmetric about y-axis.

Then, P (-x) =P(x)

P(-x)=(-x)^4+a(-x)^3+b(-x)^2+1

P(-x)=x^4-ax^3+bx+1----(4)

From (3) and (4)

x^4+ ax^3+ bx^2+ 1=x^4-ax^3+bx+1

ax^3=-ax^3

2ax^3=0

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2a=0 or x^3=0

a=0 or x=0

Therefore, a=0

Then, (3) becomes,

P(x)= x^4+ bx^2+ 1---(5)

Differentiating (5) with respect to x, we get,

P^' (x)=4x^3+2bx

Given: P(x) has a Absolute minimum at (q,-3). Then, at x=q, P^' (x)=0

0=4q^3+2bq

2q(b+2q^2 )=0

2q=0 orb+2q^2=0

q=0 or b=-2q^2

q=0 is not possible.

Therefore, b=-2q^2

P(x) passes through the point (q,-3)

Plug in the point (q,-3) in equation (5), we get,

-3= q^4+ bq^2+ 1

Plug in b=-2q^2 in above equation.

-3=q^4+(-2q^2 ) q^2+1

-3=q^4-2q^4+1

-3=-q^4+1

-q^4=-4

q^4=4

q=4^(1/4)

q=2

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Then, b=-2(2)^2

b=-2(2)

b=-4

Therefore, the required function is

P(x)=x^4-4x^2+ 1

b) Since, the graph of the symmetric about y-axis. Then, the possible values of q are 2and-2.

6) i) f ''(x) = 1/ (x+5)1/3 for all x in the domain of f

ii) the line tangent to the graph of f at (4,2) has an angle of inclination of 45 degress.

Hi,First of all the second property contains two pieces of information. The fact that (4,2) ison the graph of the function implies that f(4) = 2. Second a line with angle of inclination45o has a slope of 1. Hence f '(4) = 1.

The first property says that f ''(x) = (x+5)-1/3. Thus by antidifferentiation,

f '(x) =(3/2) (x+5)2/3 + C,

where C is a constant.

Use the fact that f '(4) = 1 to find C.

Antidifferentiate again to find an expression for f(x) and then use the fact that f(4) = 2 toevaluate the constand that arises in the andoidifferentiation.

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