Pc Functions Basic Function Problems

Solved Problems on Basic Functions

Absolute Value Problem 1

Solve the equation 2 x  1  x  5  3.

Solution Strip first the absolute value signs. Start by observerving



1 2 x  1 if x    x  5 if x  5 2 that 2 x  1   and x  5    5  x if x  5  1  2 x if x  1  2  

1 x  6 if x  2 



1 2 x  1  x  5   4  3 x if  5  x  2  6  x if x  5   

Absolute Value Problem 1

Solve the equation 2 x  1  x  5  3.

Solution (cont’d)

1 x  6  3  x  9. This value satisfies the condition x  . 2 Hence x  9 is a solution. 7 1 4  3 x  3  x   . This solution satisfies  5  x  . 3 2 Hence it is a solution. 6  x  3  x  3. This does not satisfy the condition x  5. Hence it is not a solution.

Two solutions: x = 9 and x = -7/3

Absolute Value Problem 2

Simplify the expression x  y  x  y .

Solution

If x  y ,

x  y  x  y , and

x  y  x  y  x  y   x  y   2 x. If x  y ,

x  y  y  x, and

x  y  x  y  x  y   y  x   2y . Hence  2 x if x  y xy  xy  i.e.  2y if x  y x  y  x  y  2max  x, y  .

Absolute Value Problem 3 Solution

Sketch the graph of the function f  x   x 2  4 x  3 .

If x  0, x  x. Otherwise x   x. Hence  x 2  4 x  3 if x  0 

f  x   x2  4 x  3  

 x  4 x  3 if x  0 2

.

Next observe that x 2  4 x  3  0  x  1 or x  3. Hence x 2  4 x  3  0 if 1  x  3 and positive otherwise. Next observe that x 2  4 x  3  0  x  1 or x  3. Hence x 2  4 x  3  0 if  3  x  1 and positive otherwise.

Absolute Value Problem 3

Sketch the graph of the function f  x   x 2  4 x  3 .

Solution (cont’d)

By the previous considerations  x 2  4 x  3 if x  0 

f  x   x2  4 x  3  

2  x  4 x  3 if x  0

 

x 2  4 x  3 if 0  x  1 or x  3

2   x  4 x  3 if 1  x  3  2  x  4 x  3 if  1  x  0 or x  3   x 2  4 x  3 if  3  x  1

Graph of the function f.

Absolute Value Problem 4 Solution

Sketch the graph of the equation x  x  y  y . If x  0 and y  0, x  x  y  y  2 x  2y  y  x.

If x  0 and y  0, x  x  y  y  0  2y  y  0. If x  0 and y  0, x  x  y  y  2 x  0  x  0. If x  0 and y  0, x  x  y  y  0  0. Hence all points

 x, y  , x  0 and y  0,

satisfy the equation.

Graph of the equation

Inverse Functions Problem 1 Solution

y

Find a formula for the inverse of the function f  x  

4x  1 . x 1

4x  1 Solve x in terms of y from the equation y  . x 1

4x  1 y 1  y  x  1  4 x  1   4  y  x  y  1  x  x 1 4y

This computation assumes that x  1 and y  4.

Hence f 1  x  

x 1 . 4x

The above expression for x in terms of y defines the inverse function.

It is usual to call the variable of a function by “x”. That is why one usually replaces “y” by “x” in the final answer.

Inverse Functions Problem 2

1 ex Find a formula for the inverse of the function f  x   . 1 ex

Solution

1 ex x x x y  y 1  e  1  e  e  y  1  1  y   x 1 e

 ex 

 1 y  1 y  x  ln   provided that  1  y  1. 1 y  1 y 

 1 x  Hence the inverse function is f  x   ln  .   1 x  1

Inverse Functions Problem 3 Solution

1 ex Find a formula for the inverse of the function f  x   . 1 ex 1  ex Solve x in terms of y from the equation y  . 1 ex

1  ex x x y  y 1  e  1  e    x 1 e  1 y  1 y x x 1  y e  1  y  e   x  ln    . 1 y  1 y 

This computation assumes that

 1 x  Hence f  x   ln  .  1 x  1

1 y  0. 1 y

Inverse Functions Problem 4

Which of the formulae a)

 f+g

b)

 f og  f 1og1 1  f og  g1of 1

c)

1

 f 1  g1

1

are correct? Answer

Only formula c) is correct.

Inverse Functions Problem 5

Show that the number log2 3 is irrational.

Solution

Assume the contrary, i.e., assume that log2 3 

m for some m, n  ¥ . n

m m log2 3   2 n  3  2m  3n. n

This is not possible, since if m  0 and n  0, 2m is even and 3n is odd.

Hence we cannot find integers m and n such that log2 3 

m . n

Inverse Functions Problem 6

Answer

Assume that the function f has an inverse function denoted by g. Are the following true of false: 2.

If f is increasing, then g is also increasing.

3.

If f is decreasing, then g is increasing.

4.

The function f is injective.

5.

The function g is onto.

6.

f=1/g.

7.

f+g=0.

1 is true, 2 is false, 3 is true, 4 is true, 5 is false, 6 is false.

Inverse Functions Problem 7 Solution

Simplify cos  arcsin  t   . Use the formula cos      1  sin2    .

cos  arcsin  t    1  sin2  arcsin  t    1  t 2

By the definition of the arcsin function, -π/2 ≤ arcsin(t) ≤ π/2 for all t. Hence cos(arcsin(t)) is always non-negative, and we have plus sign in the front of the square root in the answer.

Inverse Functions 



Show that f  x   ln x  x 2  1

Problem 8

is an odd function.

Solution







since



x 1 x 2



 x  x 1

 



2   ln x  x 1  f  x 

2

x 1 x  2





1

f   x   ln  x  x  1  ln  2



2

x  1  x 2  1. 2

Inverse Functions Problem 9

Solve the equation arcsin  x   arcsin  y  .

Solution

arcsin  x   arcsin  y   sin  arcsin  x    sin  arcsin  y    x  y

Inverse Functions Problem 10

Solution

Show that the inverse function of an odd function is odd.

Assume that f is odd and has an inverse function f 1.

By the definition of the inverse function, the function f is onto and one-to-one. We have to show that y : f 1   y    f 1  y  . Let y be given. Since f is onto, x such that y  f  x   x  f 1  y  . Since f is odd,  f  x   f   x  .

f 1   y   f 1   f  x    f 1  f   x     x   f 1  y  . Hence f is odd.