## Some answers to the review problems for the final exam in Math 151:04-06, fall 2003

Solutions
1 2 3 4d,e 7 10 11 12
HINT
ONLY
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.```
Submitted by
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).
```
Submitted by
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)).
```
Submitted by
Michael Pandolfo

```            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
```
Submitted by
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

```
Submitted by
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__|

```
Submitted by
Zeeshan Farman
dA/dx=1*(450-5x)/2+x*(-5x)/2=(450-10x)/2. The maximum has been found because of any one of the following reasons:
• 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)
```
Submitted by
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

```
Submitted by
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.
```
Submitted by
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
```
Submitted by
Matthew Daubert