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1 |
2 |
3 |
4d,e |
7 |
10 |
11 |
12HINTONLY |
18 |
22 |

**Problem 1**

For what values of B is the following function continuous
at x=2?

f(x)=3x^3-x^2+Bx if x>1 and is Bx-2 if x<= 1.

**Solution**

f(x) =3x^3 - x^2 + Bx if x > 1 =Bx - 2 if x <or= 1 So lim 3x^3 - x^2 + Bx = 2 + B x->1+ and lim Bx - 2 = B - 2 x->1- Set the limits equal to each other and solve for B. Then 2 + B = B - 2 So: There are no values for B that will make f(x) continuous.

Michael Pandolfo

**Problem 2**

Find the equation of any tangent lines to the curve
y=(x-1)/(x+1) that are parallel to the line x-2y=2.

**Solution**

y = (x-1)/(x+1) parallel to x - 2y = 2 but then y = (1/2)x - 1 so the slope of the line is 1/2. y' = [(1)(x+1) - (1)(x-1)] / (x+1)^2 so y' = 2/(x^2 + 2x +1). now set y' = to the slope to find x: 2/(x^2 + 2x +1) = 1/2 means (check this!) x = 1, -3. Now plug the x values into the original function to find y: the two points are (1,0) and (-3,2), so the two equations are y - 0 = (1/2)(x - 1) and y - 2 = (1/2)(x + 3).

Michael Pandolfo

**Problem 3**

Find the line tangent to the curve defined by the equation
ln(xy)+2x-y+1=0 at the point (1/2, 2)

**Solution**

Use implicit differentiation to find y': (y + xy')/xy + 2 - y' = 0 so y' = (-2xy - y)/(x - xy). plug the x and y values of the point for x and y in y' and solve for y' to get the slope of the tangent line. The slope equals 8 so the equation of the tangent line is y-2=8(x-(1/2)).

Michael Pandolfo

**Problem 4**

d)

3e^x+4e^-x lim ---------- x->infinity 5e^x+4e^-x

**Solution**

3 4 e^-x - + -------- 3e^x+4e^-x 1 e^x 3 lim ---------- = lim ------------ = - x->infinity 5e^x+4e^-x x->infinty 5 4 e^-x 5 - + -------- 1 e^x divide top and e^-x->0 and e^x->infinity bottom by e^x as x->infinity therefore 4e^-x/e^x->0

Matthew Daubert

e)

|x-8| lim ------- x->8- x-8

**Solution**

Since x<8 when x->8-... -(x-8) lim -------- x->8- x-8 cancel out the x-8 and... lim (...) = -1

Matthew Daubert

**Problem 10**

A farmer with 450 feet of fencing wants to enclose the four sides of a
rectangular region and then divide the region into four pens with
fencing parallel to one side of the rectangle. What is the largest
possible area of the fours pens?

**Solution**

First draw a picture (below) and we get 2 equations. 1) Area = x*y 2) 5x + 2y = 450 From equation 2 its found that: 3) y=(450-5x)/2 First plug in equation 3 into equation 1 ( A=x*(450-5x)/2). Take the derivative, and then solve for x. Finding x = 45, we then plug that into equation 2. solve for y. y = 112.5. finding the largest area of the 4 pens, find the area. the answer is 5062.5. Picture: __x__ |__x__| |__x__|y y|__x__| |__x__|

Zeeshan Farman

- The domain of the function A is [0,90] (since there are 5 x's and only a total of 450 feet of fence). But A(0)=0 and A(90)=0, so the max is at A(45)=5062.5.
- A'(x) is positive for x<45 and negative for x>45 and thus A(45) must be the max value of the function.
- A''(45)=-10/2<0, so A(x) is concave down at x=45 and therefore A(x) has a max at x=45.

**Problem 7**

State the formal definition of the derivative of the
function f(x). Use the definition to calculate f'(x) for
f(x)=sqrt(3-5x).

**Solution**

Derivative of f(x) = lim f(x+h)-f(x) h-->0 ----------- (x+h) - x If f(x) = (3-5x)^(1/2) then f'(x)= lim (3-5(x+h))^(1/2)-(3-5x)^(1/2)= lim (3-5(x+h))^(1/2)-(3-5x)^(1/2) h-->0 ----------------------------- h-->0 ----------------------------- (x + h) - x h _ _ = lim (3-5(x+h))^(1/2)-(3-5x)^(1/2) | (3-5(x+h))^(1/2)+(3-5x)^(1/2) | h-->0 -----------------------------*|-------------------------------| h | (3-5(x+h))^(1/2)+(3-5x)^(1/2) | |_ _| = lim (3-5x-5h)-(3+5x) = lim -5h h-->0 ------------------------------- h-->0 -------------------------------- h[(3-5(x+h))^(1/2)+(3-5x)^(1/2)] h[(3-5(x+h))^(1/2)+(3-5x)^(1/2)] = lim -5 -5 h-->0 ----------------------------- = ------------- (3-5(x+h))^(1/2)+(3-5x)^(1/2) 2(3-5x)^(1/2)

Meghan Brundage

**Problem 11**

A ladder which is 13 feet long is leaning against a wall.
Its base begins to slide along the floor, away from the wall. When
the base is 12 feet away from the wall, the base is moving at
the rate of 5 ft/sec. How fast is the top of the ladder sliding down
the wall then? How fast is the area of the triangle formed by ladder,
wall, and floor changing at that time?

**Solution**

Given: |\ c = 13ft | \ b = 12ft | \ b' = 5ft/sec a | \ c | \ | \ |______\ b a) Find a'. First find a using Pythagorean theorem: a^2+b^2=c^2 a=sqrt(13^2-b^2) a=5 Now find the derivative of a using the chain rule: a' = (1/2)[(13^2-b^2)^(-1/2)](-2b)b' And plug-in b and b': a' = (1/2)[25^(-1/2)](-24)5 a' = -12 The top of the ladder is moving towards the floor at 6ft/sec b) Find A' where A is the area of the triangle. Define the area: A = (ab)/2 Find the derivative of A using the quotient rule and product rule: (ab)'(2)-(ab)(2)' 2(a'b+ab') A' = ------------------- = ------------ 2^2 4 Substitute with known values of a, b, a', and b': 2[(-12)(12)+(5)(5)] A' = -------------------- = -119/2=-59.5 4 The area of the triangle is decreasing at a rate of 48.5ft^2/sec

Matthew Daubert

**Problem 12**

Let f(x)=3x/(x^2-1). Find the function's domain,
the intervals where f(x) is increasing or decreasing, any maxima and
minima, concavity and inflection points, and any horizontal and
vertical asymptotes of the graph of f(x). Then sketch the graph of
f(x).

**Solution**
Here's a major hint: check your answer(s) with a picture
produced by the `Maple` command

`plot(3*x/(x^2-1),x=-5..5,y=-10..10,discont=true);`
The `discont=true` allows `Maple` to skip connecting the
dots when they are far apart.
THE PICTURE IS LINKED HERE.

**Submitted by**

No one yet!

**Problem 18**

Let f(x)=3x^7 - 2x^2 + x -1. Show that f(x) must have a real
root in [0,1].

**Solution**

We know from the start that the function f(x) is continous on the=20 interval [0,1]. Because the interval is [0,1], first you plug in 0 into f(x), and find it to be -1. Then plug in 1 for f(x) and find that to be 1. Because f(x) is continous on [0,1] and moves from -1, to 1, we know that f(x) has at least one root. This process is guaranteed by the Intermediate Value Theorem.

Zeeshan Farman

**Problem 22**

Find the absolute maximum and the absolute minimum of f(x)=x/(x^2+1)
on [0,2].

**Solution**

f'(x) = (-x^2+1)/(x^2+1)^2 = 0 when -x^2+1 = 0. x=+/-1. Trash -1, it is not in the domain. Evaluate: f(0) = 0 f(1) = 1/2 f(2) = 2/5 Absolute max value (when x=1) is 1/2 Absolute min value (when x=0) is 0

Matthew Daubert