Math 1310
Review Session for the Second Midterm
1. (Linear Approximations)
√
Find a linear approximation
of the function x near 16 and use it to give an estimate
√
of the number 17. In what interval is the linear approximation around 16 giving an
error of at most 0.01?
Solution:
√
The derivative of f (x) = x is 2√1 x . Hence the linear approximation around 16 is
√
√
given by L16 (x) = 16+ 2√116 x = 4+ x−16
.
Now
we
want
to
compare
x and 4+ x−16
8
8
and we want an error of at most 0.01. Hence we set
√
x − 16
| x−4−
| ≤ 0.01
8
(1)
Hence we are interested in solving the equations
√
x − 16
= 0.01
x−4−
8
and
√
x−4−
x − 16
= −0.01
8
(2)
(3)
The first one is equivalent to
x=
2
x − 16
+ 4.01
8
privided that x ≥ 0, and the second one is equivalent to
2
x − 16
x=
+ 3.99
8
(4)
(5)
provided that x ≥ 0. Hence we can solve these quadratic equations and get the
interval we are interested in.
2. (Limits with indeterminate forms)
Calculte the following limits
(a) limx→0
sin(x)−x
;
x3
3
(b) limx→0+ x ln(x);
(c) limx→+∞ 1 +
n mx
x
(d) limx→0 tan(x)−x
x3
(e) limx→0 sin(x) ln(x)
Solution: For all of the following we can use l’Hospital’s rule.
(a)
sin(x) − x
cos(x) − 1
=
lim
x→0
x→0
x3
3x2
− sin(x)
= lim
x→0
6x
− cos(x)
= lim
x→0
6
1
=−
6
lim
(6)
(b)
ln(x)
x→0
x−3
x−1
= lim+
x→0 −3x−4
x3
= lim+ −
x→0
3
=0
lim+ x3 ln(x) = lim+
x→0
(7)
(c)
lim
x→+∞
1+
n mx
n mx
= lim eln(1+ x )
x→+∞
x
n
= lim emx ln(1+ x )
(8)
x→+∞
Since the exponential is a continuous function, it will be enough to study its
exponent.
ln 1 + nx
n
lim mx ln( 1 +
) = lim
1
x→+∞
x→+∞
x
mx
=
1 −n
1+ n
x2
x
lim
−1
x→+∞
mx2
= mn lim
x→+∞
= mn.
Page 2
1
1+
(9)
n
x
Hence we have
lim
x→+∞
1+
n mx
= emn
x
(10)
(d)
tan(x) − x
1 tan2 (x) − 1
=
lim
x→0
x→0
x3
3x2
2
sin (x)
1
lim
2
x→0
x
3 cos2 (x)
lim
Since we know limx→0
that
sin(x)
x
(11)
= 1, we can use the properties of limits to conclude
sin2 (x)
1
1
1
=1· =
2
2
x→0
x
3 cos (x)
3
3
lim
(12)
(e)
lim sin(x) ln(x) = lim
x→0
ln(x)
1
sin(x)
1
x
lim
x→0 − cos(x)
sin2 (x)
x→0
=
(13)
sin(x) sin(x)
= lim −
x→0
x cos(x)
= −1 · 0
=0
3. (Miscellaneous)
(a) Use implicit differentaition to find the tangent line to the curve given by
x2 + 2xy − y 2 + x = 2
at the point (1,2).
(b) Find the global maximum on the interval [−2, 2] of the function f (x) = x2 − x + 1.
(c) Using first and second derivative, sketch the graph of the following functions f (x) =
x3
.
2(x3 +1)
Solution:
Page 3
(a) Using implicit differentiation we get 2x + 2y + 2xy 0 − 2yy 0 + 1 = 0. Plugging in
(1,2) we get 2 + 4 + 2y 0 − 4y 0 + 1 = 0, which gives us y 0 = 72 . Hence the line is
given by y − 2 = 72 (x − 1).
(b) The derivative is f 0 (x) = 2x − 1. It is zero, i.e. f 0 (x) = 0, if 2x − 1 = 0, hence
if3
1
1
1
x = 2 . So the values to be tested are the extrema and 2 . f (−2) = 7, f 2 = 4
and f (2) = 3. So the global maximum is 7 and it is reached at -2.
4. (Optimization and Derivatives)
(a) Find an equation of the line through the point (3,5) that cuts off the least area from
the first quadrant.
Solution:
First we notice that to cut a bounded region in the first quadrant the slope m
has to be negative. Hence from now on m < 0. Now the equation of a line
(neither vertical nor horizontal) through (3,5) is y − 5 = m(x − 3). Then we
want to find where this line crosses the x-axis and the y-axis. If y = 0, we get
x = 3m−5
, while for x = 0 we get y = 5 − 3m. Then the points are ( 3m−5
, 0)
m
m
1
3m−5
1 (3m−5)2
and (0, 5 − 3m). The area described is hence 2 (5 − 3m) m = − 2 m . To
find the minimum we need to find the critical values of the derivative. Hence
we end up with the equation
−
1 2(3m − 5)3m − (3m − 5)2
=0
2
m2
(14)
which, after clearing denominators, leads to
(3m − 5)(6m − 3m + 5) = (3m − 5)(3m + 5) = 0
(15)
Since m < 0, the solution is given by 3m + 5 = 0. Hence we have just one
critical point. Looking at the picture, we get it has to be the minimum, which
is hence unique. Thus we get m = − 35 .
(b) What is the shortest possible length of the line segment that is cut off by the first
quadrant and is tangent to the curve y = x3 at some point?
Solution: We want to see what is the segment cut off by the tangent line to
the point (x0 , x30 ). We have that y 0 = − x32 and hence the line we are looking for
has equation
3
3
y−
= − 2 (x − x0 )
(16)
x0
x0
Page 4
If we set x = 0, we get
y−
3
3
=
x0
x0
6
y=
x0
(17)
while if we set y = 0, we get
−
3
3
= − 2 (x − x0 )
x0
x0
1
1 = (x − x0 )
x0
x = 2x0
(18)
Now, we have to consider the segment joining (2x0 , 0) and (0, x60 ). The points
of minimum of the distance d are the same as the ones of its square, d2 , and
hence we can look for the minimum points of 4x20 + x362 . Hence we look for the
0
critical values of its derivative
(d2 (x0 ))0 = 8x0 −
72
x30
(19)
If we set (d2 (x0 ))0 = 0 and we clear the denominators, we get
8x40 − 72 = 0
8x4 = 72
x4 = 9
√
x= 3
Page 5
(20)