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Math 171 Review for the Final Exam Solutions
5. Find the values of a and b , if the tangent line to y =
ax^{2}  bx at
(2, 5) has slope m = 2. (5 points)
Solution. Because y(2) = 5 we have the following equation for a
and b.
4a + 2b = 5
The derivative is 2ax  b. Because we have the second
equation for a and b .
4a  b = 2.
Adding the equations we get
b = 7
and plugging this value of b into the first equation we get 4a + 14 = 5
whence
6. If a ball is thrown vertically upward with a velocity of 80 ft/s, then its
height after t seconds is h(t) = 80t  16t^{2}. What is the velocity of the
ball when it is 96 ft above the ground on its way up? On its way
down? (5 points)
Solution The velocity of the ball equals to the derivative of its height.
We can find the moments of time when the ball is 96 ft high
by
solving the quadratic equation
80t  16t^{2} = 96.
Dividing both parts by 16 and moving all terms to the left we get
t^{2}  5t + 6 = 0
whence t = 2 or t = 3. At the moment t = 2 the velocity of the ball is
16 ft/sec and the ball is moving up whilst at the moment t = 3 the
velocity is 16 ft/sec and the ball is moving down.
7. An open box is to be made from a 6ft by 8ft rectangular piece of
sheet metal by cutting out squares of equal size from the four corners
and bending up the sides. Find the maximum value that the volume
of the box can have.(10 points)
Solution Let x be the side of each cut out square. Then the height of
the box is x
whilst its length is 82x and its width is 62x (See the picture below).
Notice that because these dimensions cannot be negative we
have
0 ≤ x ≤ 3. There fore we have to maximize the volume of the box
V (x) = x(82x)(62x) = 4x(4x)(3x) = 4x(x^{2}7x+12) = 4(x^{3}7x^{2}+12x)
on the interval [0, 3]. At the ends of the interval the function V takes
value 0 so it will take the greatest value at a stationary point inside
the interval. To find the stationary points we have to solve the
equation
The quadratic formula provides two solutions
The sign + provides a solution which is greater than 3 and
of no
interest to us. The only stationary point inside the interval [0, 3] is
You can easily check yourself that the greatest value of the volume is
8. Sketch the graph of y = x^{3}  3x + 2. Find x and y
intercepts, plot
the stationary points and the inflection points and determine the
intervals where y is increasing and decreasing, concave up and
concave down.(10 points)
Solution. To find the xintercepts we have to solve the equation
x^{3}  3x + 2 = 0.
Notice that x = 1 is a solution of this equation and therefore x  1 is
a factor of x ^{3}  3x + 2. We can factor x^{3}  3x + 2 using long division,
grouping , or synthetic division. The table below shows synthetic
division.
Therefore x^{3}  3x + 2 = (x  1)(x^{2} + x  2) = (x  1)^{2}(x +
2) and the
x intercepts are (2, 0) and (1, 0).
The yintercept is (0, 2).
The first derivative of y is
The function has two stationary points, 1 and 1 which
divide the
xaxis into the intervals (∞,1), (1, 1), and (1,∞). The next
table shows the behavior of the function on these intervals.
Interval  (∞,1)  (1, 1)  (1,∞) 
Sign of  +    + 
Behavior of y  Increases  Decreases  Increases 
To find the inflection points of the cubic function y we
compute its
second derivative
We see that y has only one inflection point 0. The
information about
concavity of y is contained in the next table.
Interval  (1, 0)  (0,1) 
Sign of    + 
Concavity  Down  Up 
A computer generated graph of the function y = x^{3}  3x + 2
is shown
below.
9. Use the fundamental theorem of calculus:(4 points each)
(a) Find the derivative of the function
Solution. The fundamental theorem of calculus in its
generalized form tells us that
Applying this formula when a (x) = x^{2}, b(x) = sin x, and
f(t) = sin t^{2} we get
(b) Prove that the function
is a constant on the interval (∞, 0).
Solution. Look at two ways to solve the problem.
First way. We apply the fundamental theorem of calculus to see
that
Because the derivative of F is identically 0 the function
F is a
constant function.
Second way.
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