Derivative Tests
Math 1300
Ada Chan
November 4, 2014
Math 1300 (York University)
Derivative Tests
1/1
Denition: Let f be a function whose domain contains an interval I.
We say
f is increasing on I if f (x1 )<f (x2 ) whenever x1 < x2 in I.
f is decreasing
Example 4 Find lim
0
sin 5
.
tan 8
Solution We know about limits of the sine of an angle which approaches zero,
but nothing about limits of tangents. Thus rewrite this limit in terms of sin:
lim
0
sin 5
sin 5
sin 5
= lim
= lim
cos 8 .
tan 8 0 sin 8 / co
Example 5 Find lim ( tan sec ).
2
Solution We only know limits of sine and cosine. Hence start by
rewriting this limit in terms of sine and cosine:
1
sin 1
sin
lim ( tan sec ) = lim
.
= lim
cos cos cos
2
2
2
We only know limits as an angle approaches
Example 6 Find lim ( 1) csc .
1
Solution We only know limits involving sine and cosine, so rewrite the
given limit in terms of sine and cosine:
1
lim ( 1)csc = lim ( 1)
.
1
1
sin
Also, we only know facts about trigonometric limits when the angle
approa
Example 1 Find lim ( 5x- 4).
x 3
Solution As x gets closer and closer to 3 , the values of 5x-4 get closer and
closer to 5 (3 )-4 = 11 . Alternatively, as x gets closer and closer to 3 the
point (x,5x-4 ) on the graph of y = 5x-4 gets closer and closer to
Example 3 Find lim
sin
.
Solution Note that sin is always a number between 1 and +1. We are
dividing this number by which is getting larger and larger. This quotient
gets closer and closer to zero. To give a proper justification that the limit
is zero,
Example 2 Find lim cos
1
.
Solution We know that
lim
1
= 0.
By Property 6 (the limit of a composite function is the
value of the second function applied to the limit of the
first function):
lim cos
1
1
= cos lim = cos 0 = 1.
1
Example 4 Solve the initial value problem
dy
1
=
,
dx 1 + x 2
y(1) = 0.
1
Solution We know that the derivative of arctan x is
.
1+ x2
solution to the differential equation is:
Hence the
y = arctan x + C
where C is an arbitrary constant. Then
0 = y( 1) = a
|x 5|
.
x5 x 2 + 1
Example 6 Find lim
Solution Observe that
Hence for x > 5
x5
if x 5
|x 5| x 2 + 1
=
x 2 + 1 (x 5)
x 2 + 1 if x < 5
lim ( x 5)
x5
0
|x 5|
= lim 2
= x5 + 2
=
=0
2
x 5 + x + 1
x5 + x + 1
lim ( x + 1) 26
lim
while for x<5
x5+
( x 5) xlim
Example 1 Determine where the function f(x) = x3 - 3x2 - 24x + 15
is increasing and where it is decreasing.
Solution First, compute the derivative of f:
f (x ) = 3 x 2 6 x 24 = 3( x 2 2 x 8) = 3(x 4)( x + 2).
Hence f /(x) is positive for x<-2 or x>4 while
Example 7 Find lim x sin
x 0
1
.
x
Solution In Example 1.4B2(c) we sketched the graph of y = x sin 1/x .
y
y=x
x
y = x sin 1/ x
y=-x
Note that this graph approaches the origin as x approaches zero. Thus it appears that
the value of our limit is zero. We g
Example 2 Determine where the function g( x) =
is increasing and where it is decreasing.
x+3
x2 5
Solution First, compute the derivative of g by the quotient rule:
g( x ) =
=
D(x + 3)( x 2 5) (x + 3) D( x 2 5) ( 1)(x 2 5) ( x + 3)( 2 x )
=
( x 2 5) 2
( x
Example 8 Find lim
x 25
3
x + 2.
Solution Let y = 3 x + 2 . Then y3 = x+2 a nd x = y3 -2 . Define the function
x = f (y) = y3 -2
with domain . Then x = f(y) i s a one-to-one function with inverse function
y = f 1( x ) = 3 x + 2 .
Note f (3 )=25 and thus f
Example 3 Determine where the function h(x) = |x2-4x-12|
is increasing and where it is decreasing.
Solution Note that h(x) = |(x-6)(x+2)|. Hence h(x) = x2-4x-12 for x-2 or
x6 while h(x) = -(x2-4x-12) for -2<x<6. Therefore
h/(x) = 2x-4=2(x-2) for x<-2 or x
Example 1 Find lim sin .
2
Solution We only know limits of sine and cosine as an angle approaches
zero. Therefore, we change variables in this limit from to = - /2 .
As approaches /2 , approaches 0 . Thus
lim sin = lim sin + = lim sin cos + cos sin
0
2 0
Example 4 Determine where the function k(x) = sin x + cos x with domain
[0,4] is increasing and where it is decreasing.
Solution Note that k/(x) = cos x sin x. Compare the graphs of y = cos x and
y
y = sin x.
y=cos x
4
5
4
9
4
13
4
4
x
y=sin x
k/(x)>0 and