Transcript Properties of Functions

Properties of Functions
Sketch the following function f  x   x3  x 2  x  1
x  0  f  0   1  y  intercept at  0,1
y  0  0  x3  x 2  x  1
y  0  0   x  1 x  1 x  1   x  1  x  1
2
x  1 or x  1 cuts the x  axis at 1, 0  ,  1, 0 
f ''  x   6 x  2
f '  x   3x 2  2 x  1  0
 3x  1 x  1  0
1
SP at x   x  1
3
f 
1
3
      
1 3
3
1 2
3
f ''   13   4
f ''  x   0 hence maximum SP
f '' 1  4
f ''  x   0 hence minimum SP
32
     1   max at   13 , 32
27 
27
1
3
f 1  1  1  1  1  0
3
2
 min at 1, 0 
Odd Functions
A function is said to be odd if
f   x    f  x  for every value in the
domain of x
The graph is symmetrical under 180 about
the origin
y
y  x3
y
even Functions
A function is said to be even if
f   x   f  x  for every value in the
domain of x
The graph is symmetrical under reflection
in the y  axis
y
y  x2
y
First derivative test.
1. Differentiate
2. Set derivative equal to zero
3. Use nature table to determine the behaviour of the graph
Second derivative test
1. Differentiate
2. Set derivative equal to zero
3. Find second derivative
4. Substitute x values in to second derivative
5. If second derivative is positive, minimum
6. If second derivative is negative, maximum
7. If second derivative is zero or does not exist, use
nature table
An asymptote is a line at which the rational
polynomial in the form of h  x  
f  x
g  x
is undefined
1
1
y
If x  0, y is undefined, since is not allowed
x
0
x  0is therefore a vertical asymptote
1
y  0  0  which is also impossible
x
y  0 is therefore a horizontal asymptote
1

x  0 , y    is positive 
x

1

x  , y  0 from above  is positive 
x

1


x  0 , y    is negative 
x

1

x  , y  0 from below  is negative 
x


x 3
Sketch the graph of y  2
x  x2
0  3
3
x0 y
  y  intercept is
2
 0  0  2 2
y 00
 3
 0, 
 2
x 3
 x  3  0  x  intercept is 3, 0
2
x  x2
To find any vertical asymptotes, we set the denominator
equal to zero
x2  x  2  0
 x  2  x  1  0
x  2 and x  1 are vertical asymptotes
x
y
-2.1

-2
-1.9 0.9
 
1
1.1

Non vertical Asymptotes
What happens to the y value if x tends to  infinity
x 3
For y  2
the degree of the denominator is greater
x  x2
than the numerator, hence the function tends to zero.
y  0 is a non vertical asymptote.
x 3
Sketch the graph of y  2
x  x2
0  3
3
x0 y
  y  intercept is
2
 0  0  2 2
y 00
 3
 0, 
 2
x 3
 x  3  0  x  intercept is 3, 0
2
x  x2
To find any vertical asymptotes, we set the denominator
equal to zero
x2  x  2  0
 x  2  x  1  0
x  2 and x  1 are vertical asymptotes
x
y
-2.1

-2
-1.9 0.9
 
1
1.1

Non vertical Asymptotes
What happens to the y value if x tends to  infinity
x 3
For y  2
the degree of the denominator is greater
x  x2
than the numerator, hence the function tends to zero.
y  0 is a non vertical asymptote.
A function is defined by
2x  x 1
f  x 
x 1
2
y  f  x
Find the coordinates off the points where
the graph y  f  x  crosses the coordinate axes
Find the equation of all vertical and non vertical asymptotes
Find the coordinates of any stationary points, and, if they
exist determine their nature.
Sketch the graph of y  f  x 
State the range of values of the constant k such that
the equation f  x   k has no real solution
f  0 
2  0  0 1
2
 0 1
y  intercept at  0,1
1

1
1
2x2  x 1
0
 2 x2  x 1  0
x 1
 2 x  1 x  1  0
1
x  and x  1
2
1 
x  intercept at  , 0  and  1, 0 
2 
Vertical Asymptote when
x 1  0  x  1
x 0.9 1
1.1
y 

2x  3
 x  1 2 x 2  x  1
2x2  2x
3x  1
3x  3
2
2
y  2x  3 
x 1
x   y  2 x  3
y  2 x  3 is a non vertical asymptote
x   y  
x   y  
2
1
y  2x  3 
 2 x  3  2  x  1
x 1
dy
2
2
 2  2  x  1  2 
2
dx
x

1
 
0  2
2
 x  1
2
2  x  1  2  0
2
 x  1  1
x  1  1
x  1  1  2
x  1  1  0
2
2
f  2  2  2  3 
 9  SP  2,9 
  2  1
2
f  0  2  0  3 
 1  SP  0,1
  0  1
dy
2
 2  2  x  1
dx
d2y
4
3

4
x

1

 
3
2
dx
 x  1
d2y
4
x  0: 2 
 4   0,1 max
3
dx
  0   1
d2y
4
x  2: 2 
 4   2,9  min
3
dx
  2   1