New functions from old function

NEW FUNCTIONS FROM OLD FUNCTIONS

In this section we start with the basic functions we discussed and obtain new functions by shifting, stretching, and reflecting their graphs. We also show how to combine pairs of functions by the standard arithmetic operations and by composition.

 

1. COMBINATIONS OF FUNCTIONS

 

Two functions ƒ and g can be combined to form new functions f+g,f‒g. fg. and f/g in a manner similar to the way we add, subtract multiply, and divide real numbers. The sum and difference functions are defined by

(f + g)(x) = f(x) + g(x)

(fg)(x) = f(x) − g(x)

If the domain of ƒ is A and the domain of g is B, then the domain of ƒ + g is the intersection A∩B because both f(x) and g(x) have to be defined. For example, the domain of f(x) = √x is A = [0, ∞) and the domain of g(x) = √[2‒x] is B = (‒∞, 2]. So the domain of (f + g)(x) = √x + √[2‒x] is A∩B = [0,2].

Similarly, the quotient functions are defined by

 (fg)(x) = f(x)g(x)

(f/g) (x) = f(x) / g(x)

The domain of fg is A∩B, but we can't divide by 0 and so the domain of f/g is {x ϵ A∩B| g(x) ≠ 0). For instance, if f(x) = x² and g(x) = x‒1, then the domain of the rational function (f/g)(x) = x²/(x‒1) is {x|x ≠ 1}, or (‒∞,1) U (1, ∞).

 

 

PROBLEMS UNDER COMBINATION OF FUNCTIONS

 

Example 14. Let f(x) = 1 +√[x‒2] and g(x) = x‒3. Find the domains and formulas for the functions i) f+g, ii) ƒg, iii) f‒g, iv) f/g, and v) 7f.

Solution: First, we will find the formulas and then the domains.

i) (f + g)(x) = f(x) + g(x)

= (1 + √[x‒2])+(x − 3)

= x ‒ 2 + √(x‒2

ii) (ƒ‒g)(x) = f(x) ‒ g(x)

 = (1 + √(x – 2)) ‒ (x − 3)

= 4‒x + √[x‒2])

iii) (fg)(x) = f(x)g(x)

 = (1 + √[x‒2])(x‒3)

iv) (f/g)(x) = f(x)/g (x) = { 1+√[x‒2] } / {x‒3}

v) (7ƒ)(x) = 7f(x) = 7 + 7√(x‒2)

The domains of ƒ and g are [2, +∞) and (‒∞, +∞), respectively (their natural domains). Thus, it follows from the definition, namely,

[2, + ∞) ∩ (‒∞, +∞) = [2, +∞)

Moreover, since g(x) = 0 if x = 3, the domain of f/g is [2, +∞) with x = 3 removed, namely,

[2,3) U (3, +∞)

Finally, the domain of 7f is the same as the domain of f.

 

Example 15. Show that if f(x) = √x. g(x) = √x, and h(x) = x, then the domain of f g is not the same as the natural domain of h.

Solution: The natural domain of h(x) = x is (‒∞, +∞).

Note that (fg)(x) = √x√x = x = h(x) on the domain of fg.

The domains of both ƒ and g are [0, +∞), so the domain of f g is

[0, +∞) ∩ [0, +∞0) = [0, +∞)

Since the domains of fg and h are different, it would be misleading to write (fg)(x) = x without including the restriction that this formula holds only for x ≥ 0.

 

Example 16. For the graphs of y = √x and y = 1/x, make a sketch that shows the general shape of the graph of y = √x+1/x for x≥ 0.

Solution:

To add the corresponding y‒values of y = √x and y = 1/x graphically, just imagine them to be "stacked" on top of one another.

 

 

2. COMPOSITION OF FUNCTIONS

 

In general, given any two functions f and g, we start with a number x in the domain of g and find its image g(x). If this number g(x) is in the domain of f, then we can calculate the value of f (g(x)). The result is a new function h(x) = f(g(x)) obtained by substituting g into f. It is called the composition (or composite) of f and g and is denoted by f^og ("f circle g").

Definition: Given two functions f and g, the composite function f ^og (also called the composition of ƒ and g) is defined by

(f^og)(x) = f(g(x))

The domain of f ^og is the set of all x in the domain of g such that g(x) is in the domain of f. In other words, (f ^og) (x) is defined whenever both g(x) and f(g(x)) are defined. Figure 11 shows how to picture f ^o g in terms of machines.

It is possible to take the composition of three or more functions. For instance, the composite function f ^og^oh is found by first applying h, then g, and then ƒ as follows:

 (f°g ^o h)(x) = ƒ (g(h(x)))

 

 

PROBLEMS UNDER COMPOSITION OF FUNCTIONS

 

Example 17. If f(x) = x2 and g(x) = x − 3 find the composite functions fog and gof.

Solution:

We have

(f ^og)(x) = f(g(x)) = f(x − 3) = (x − 3)²

 (g ^o f)(x) = g(f(x)) = g(x²) = x² − 3

NOTE: You can see from above Example that, in general, f^og ≠ g^of. Remember, the notation f^og means that the function g is applied first and then ƒ is applied second. In the above Example, f^og is the function that first subtracts 3 and then squares; g^of is the function that first squares and then subtracts 3.

 

Example 18. Let f(x) = x² + 3 and g(x) = √x. Find

(a) (f ^og)(x)

(b) (g ^o f)(x)

Solution: (a). The formula for f(g(x)) is

ƒ(g(x)) = [g(x)]² + 3 = (√x)² + 3 = x + 3.

Since the domain of g is [0, +∞) and the domain of ƒ is (‒∞, ∞), the domain of f^og consists of all x in [0, +∞) such that g(x) = √x lies in (‒∞, +∞); thus, the domain of fog is [0, +∞). Therefore,

(f ^og) (x) = x + 3, x≥0

(b). The formula for g(f(x)) is

g(f(x)) = √f(x) = √[x² + 3]

Since the domain of f is (‒∞, +∞) and the domain of g is [0, +∞), the domain of g^of consists of all x in (‒∞, +∞) such that f(x) = x²+ 3 lies in [0, +∞). Thus, the domain of g^of is (‒∞, +∞). Therefore,

(g ^o f)(x) = √[x²+3]

There is no need to indicate that the domain is (‒∞, +∞), since this is the natural domain of √[x² + 3].

 

Example 19. Express sin(x³) a composition of two functions

Solution: To evaluate sin(x³) we would first compute x³ and then take the sine, so g(x) = x³ is the inside function and f(x) = sin x the outside function.

Therefore, sin(x³) = f(g(x))

 

Example 20. If f(x) = √x and g(x) = √[2‒x], find each function and its domain.

(a) f^og (b) g^of (c) f^of (d) g^og

Solution:

(a) (f^og) (x) = f(g(x)) = f(√[2 –x])) = √√[2‒x] = ⁴√[2‒x]

The domain of f^og is { x|2‒x ≥ 0} = {x|x ≤ 2} = (‒∞,2].

(b) (g ° f) (x) = g(f(x)) = g(√x)) = √√[2 ‒ √x]

For √x to be defined we must have x ≥0. For √[2‒√x] to be defined we must have 2‒√x ≥ 0, that is, √x ≤ 2 or x ≤ 4. Thus 0 ≤ x ≤ 4, so the domain of g ^o f is the closed interval [0,4].

(c) (f°f) (x) = f(f(x)) = ƒ(√x) = √√x = ⁴√x

The domain of f ^o f is [0, ∞).

(d) (g ^o g) (x) = g(g(x)) = g(√[2 –x]) = √[2 ‒ √[2‒x] ]

This expression is defined when both 2‒x ≥ 0 and 2 ‒ √[2‒x] ≥ 0. The first inequality means x ≤ 2, and the second is equivalent to √[2‒x] ≤ 2, or 2‒x ≤ 4 or x ≥ − 2. Thus ‒2 ≤ x ≤ 2 so the domain of g^og is the closed interval [−2,2].

 

Example 21. Find (f^og^oh)(x) if

f(x)=√x, g(x) = 1/x, h(x) = x³

Solution:

(f ^og^oh)(x) = f(g(h(x))) = f(g(x³))

 

Example 22. Find f o g o h if f(x) = x / (x + 1). g(x) = x¹⁰. and h(x)= x+3.

Solution: (f^og^oh)(x) = f(g(h(x))) = f(g(x+3))

= f((x+3)¹⁰) = [ (x+3)¹⁰ ] / [ (x+3)¹⁰ + 1 ]

So far we have used composition to build complicated functions from simpler ones. But in calculus it is often useful to be able to decompose a complicated function into simpler ones, as in the following example.

 

Example 23. Given F(x) = cos²(x+9) find functions f, g, and h such that F = f^og^oh.

Solution: Since F(x)= [cos(x+9)]² the formula for F says: First add 9, then take the cosine of the result, and finally square. So we let

h(x) = x+9

g(x) = cos x

f(x) = x²

Then, (f o g o h)(x) = f (g(h(x))) = f(g(x + 9)) = f(cos(x + 9))

 [cos(x+9)]² = F(x)

 

Example 24. Use the table to evaluate each expression.

(a) f(g(1)) (b) g(f(1)) (c) f(f(1)) (d)g(g(1)) (e) (gof)(3) (f) (fog)(6)

Solution:

a) f(g(1)) = ƒ(6) = 5

b) g(f(1)) = g(3) = 2

c) ƒ(ƒ(1)) = ƒ(3) = 4

d) g(g(1)) = g(6) = 3

e) (gof)(3) = g(f(3)) = g(4) = 1

f) (fog)(6) = f(g(6)) = f(3) = 4

 

 

EXERCISE

 

3. Find f+g,f‒g.fg and f/g and state their domains.

(a) f(x) = x³ + 2x², g(x) = 3x²‒1

(b) f(x)=√[3‒x], g(x) = √[x²‒1]

 

4. Find the functions (a) fog. (b) gof (c) f of and (d) g o g and their domains.

(a) f(x) = x² ‒ 1, g(x) = 2x + 1

(b) f(x) = x ‒2, g(x) = x² + 3x + 4

(c) f(x) = 1‒3x, g(x) = cos x

(d) f(x) = √x, g(x) = ³√[1‒x]

(e) f(x) = x + 1/x, g(x) = x+1 / x+2

(f) f(x) = x / 1+x , g(x) = sin 2x

 

5. Find f^og^oh.

(a) f(x) = x+1, g(x) = 2x, h(x) = x ‒ 1

(b) f(x) = 2x − 1, g(x) = x², h(x) = 1 ‒ x

(c) f(x) = √[x‒3], g(x) = x², h(x) = x² + 2

(d) f(x) = tanx, g(x) = x / x‒1, h(x) = ³√x

 

6. Express the function in the form f^og.

(a) F(x) = (x² + 1)¹⁰

(b) F(x) = sin(√x)

(c) F(x) = ³√x / 1+³√x

(d) G(x) = ³√ [ x / 1+x ]

(e) u(t) = √cost

(f) u(t) = tant / 1+tant

 

7. Express the function in the form f^og^oh.

(a) H(x) = 1‒3^x2

(b) H(x) = ⁸√[ 2 + |x| ]

(c) H(x) = sec⁴(√x)

 

 

3. ODD AND EVEN FUNCTIONS

 

(a) f(‒x) = (‒x)²

= x² = f(x)

Therefore f is an even function.

(b) f(‒x) = |‒x|

= |x| = f(x)

Therefore f is an even function.

(c) f(‒x) = cos(‒x)

= cos x = f(x)

Therefore f is an even function.

(d) f(‒x) = ‒x = ‒f(x)

Therefore f is an odd function.

(e) f(‒x)=sin(‒x)

= ‒sin x = ‒ f(x)

Therefore f is an odd function.

 

Example 26. Determine whether each of the following functions is even,

odd, or neither even or odd.

(i) f(x) = x sinx

(ii) f(x) = xcosx

(iii) f(x) = x + x³

(iv) f(x) = 1‒x⁴

(v) f(x) = x⁵+x

(vi) f(x) = 2x ‒ x²

Solution:

(i) f(x) = ‒x sin (‒x)

= ‒x (‒ sin x) = x sinx = f(x)

Therefore f is an even function.

(ii) f(‒x) = ‒x cos (‒x)

= ‒x cos x = f(x)

Therefore f is an odd function.

(iii) f(‒x) = ‒x+(‒x)³

= ‒x‒x³ = (x + x³) = ‒f(x)

Therefore f is an odd function.

(iv) f(‒x) = 1− (−x)⁴

= 1− x⁴ = f(x)

Therefore f is an even function.

(v)ƒ(−x) = (‒x)⁵ ‒ x

= −x⁵ − x = −(x⁵+x) = −f(x)

Therefore f is an odd function.

(vi) f(−x) = 2(‒x) ‒ (‒x)²

= ‒2x ‒ x² = −(2x + x²)

Since f(‒x) ≠ f(x) and f(‒x) ≠ ‒f(x), we conclude that f is neither even nor odd.

 

 

EXERCISE

 

8. Determine whether each of the following functions is even, odd, or neither even

or odd.

(a) f(x) = x² + 1        Ans: f is an even function

(b) ƒ(x) = x³ + x       Ans: f is an odd function

(c) ƒ (x) = 1 / x²‒1      Ans: f is an even function

(d) f(x) = 1 / x‒1     Ans: f is neither odd nor even

(e) f(x) = 2x+1       Ans: f is neither odd nor even

 

 

4. TRANSFORMATIONS OF FUNCTIONS

 

By applying certain transformations to the graph of a given function we can obtain the graphs of certain related functions. This will give us the ability to sketch the graphs of many functions quickly by hand. It will also enable us to write equations for given graphs.

 

TRANSLATION

Now let's consider the translation transformations. If c is a positive number. then the graph of y = f(x)+c is just the graph of y = f(x) shifted upward a distance of c units (because each y‒coordinate is increased by the number c).

By the similar way, the graph of y = f(x) ‒ c is just the graph of y = f(x) shifted downward a distance of c units (because each y‒coordinate is decreased by the number c).

The graph of y = f(x + c), where c> 0, then is just the graph of y = f(x) shifted c units to the left. By the similar way, the graph of y = f(x‒ c) is just the graph of y = f(x) shifted c units to the right.

 

VERTICAL AND HORIZONTAL SHIFTS

Suppose c>0 To obtain the graph of

• y = f(x) + c shift the graph of y = f(x) a distance c units upward

• y = f(x) c shift the graph of y = f(x) a distance c units downward

• y = f(x + c) shift the graph of y = f(x) distance c units to the left.

• y= f(x‒c) shift the graph of y = f(x) distance c units to the right

 

STRETCHING, COMPRESSION AND REFLECTING

Now let's consider the stretching and reflecting transformations. If c > 1 then the graph of y = cf(x) is the graph of y = f(x) stretched by a factor of c in the vertical direction (because each y‒coordinate multiplied by the same number c). The graph of y = f(x) is the graph of y = f(x) reflected about the x‒axis because the point (x, y) is replaced by the point (x,y) (the following chart, where the results of other stretching, compressing, and reflecting transformations are also given.)

 

VERTICAL AND HORIZONTAL STRETCHING AND COMPRESSION

To obtain the graph of

• y = cf(x) stretch the graph of y = f(x) vertically by a factor of c

• y = 1/c f(x) compress the graph of y = f(x) vertically by a factor of c

• y = f(cx) compress the graph of y = f(x) horizontally by a factor of c

• y = f(x/c) stretch the graph of y = f(x) horizontally by a factor of c

 

REFLECTION

• y = −f(x) reflect the graph of y = f(x) about the x‒axis

• y = ƒ(− x) reflect the graph of y = f(x) about the y‒axis

 

Example 27. Sketch the graph of the function y = | x² ‒ 1 |.

Solution: y = x²‒1 1 is a parabola with symmetrical about y‒axis with vertex (1,0). i.e., by shifting the parabola y = x² downward 1 unit. When −1 < x < 1, y = x² ‒ 1 is negative. We see that the graph lies below the x‒axis when ‒1 < x < 1. So we reflect that part of the graph about the x‒axis to obtain the graph of y = |x² ‒ 1|.

 

Example 28. Sketch the graph of (a) y=√(x‒3) (b) y = √(x+3)

Solution:

a) Using the translation principles given, the graph of the equation y = √(x‒3) can be obtained by translating the graph of y = √x right 3 units.

b) The graph of y = √(x+3) can be obtained by translating the graph of y = √x left 3 units.

 

Example 29. Sketch the graph of y = x² ‒ 4x + 5.

Solution: Completing the square on the first two terms yields

y = (x² ‒ 4x + 4) ‒ 4 + 5 = (x‒2)² + 1

In this form we see that the graph can be obtained by translating the graph of y = x² right 2 units because of the x‒2, and up 1 unit because of the +1.

 

Example 30. Given the graph of y = √x, use transformations to graph y = √x‒2. y = √(x‒2), y = ‒√x, y = 2√x, and y = √(‒x).

Solution:

Given y = √x. i.e., y² = x is a parabola passing through origin symmetrical about x‒axis. In the other parts of the figure we sketch y=√x‒2 shifting 2 units downward, y = √(x ‒ 2) by shifting 2 units to the right, y = ‒√x by reflecting about the x‒axis, y = 2√x by stretching vertically by a factor of 2, and y = √‒x reflecting about the y‒axis.

 

Example 31. Sketch the graph of y = ³√(2‒x).

Solution: Using the translation and reflection principles, we can obtain the graph by a reflection followed by a translation as follows: First reflect the graph of y = ³√x about the y‒axis to obtain the graph of y = ³√‒x, then translate this graph right 2 units to obtain the graph of the equation y = ³√[‒(x‒2)] = ³√[2‒x]

 

Example 32. Sketch the graph of y = 4 ‒ | x ‒ 2|.

Solution: The graph can be obtained by a reflection and two translations: First translate the graph of y = |x| right 2 units to obtain the graph of y = |x ‒ 2|; then reflect this graph about the x‒axis to obtain the graph of y = ‒|x‒2|; and then translate this graph up 4 units to obtain the graph of the equation y = ‒ |x − 2| + 4 = 4 ‒ |x ‒ 2|.

 

Example 33. Sketch the graphs of the following functions.

(a) y = cos 2x

(b) y = cos (1/2)x

Solution: (a) We obtain the graph of y = cos 2x from that of y = cos x by compressing horizontally by a factor of 2. Thus, whereas the period of y = cos x is 2π, the period of y = cos 2x is 2π/2 = π.

(b) We obtain the graph of y = cos ½ x from that of y = cos x by stretching horizontally by a factor of 2. Thus, whereas the period of y = cos x is 2π, the period of y = = cos ½ x is 2π × 2 = 4π.

 

Example 34. Sketch the graphs of the following functions.

(a) y = 2cos x

(b) y = 1/2 cos x

Solution: (a) We obtain the graph of y = 2 cosx from that of y = cos x by stretched vertically by a factor of 2. Thus, whereas the period of y = cos x is 2π, the period of y= 2 cos x is also 2π.

(b) We obtain the graph of y = 1/2 cosx from that of y = cos x by compress vertically by a factor of 2. Thus, whereas the period of y = cos x is 2π, the period of y = ½ cosx is also 2π.

 

Example 35. Sketch the graphs of the following functions.

(a) y = sin 2x

(b) y = 1 ‒ sin x

(a) We obtain the graph of y = sin 2x from that of y = sin x by compressing horizontally by a factor of 2. Thus, whereas the period of y = sin x is 2π, the

period of y = sin 2x is 2π/2 = π.

(b) To obtain the graph of y = 1‒ sin x, we again start with y sin x. We reflect about the x‒axis to get the graph of y = ‒ sin x and then we shift 1 unit upward to get y = 1‒ sin x.

 

 

EXERCISE

 

9. Suppose the graph of ƒ is given. Write equations for the graphs that are obtained

from the graph of ƒ as follows.

(a) Shift 3 units upward.

(b) Shift 3 units downward.

(c) Shift 3 units to the right.

(d) Shift 3 units to the left.

(e) Reflect about the x‒axis.

(f) Reflect about the y‒axis.

(g) Stretch vertically by a factor of 3.

(h) Shrink vertically by a factor of 3.

 

10. Explain how each graph is obtained from the graph of y = f(x).

(a) y = 5f(x)

(b) y = f(x‒5)

(c) y = ‒f(x)

(d) y = ‒5f(x)

(e) y = f(5x)

(f) y = 5f(x) ‒3

 

11. The graph of y = f(x) is given. Match each equation with its graph and give

reasons for your choices.

(a) y = f(x‒4)

(b) y = f(x)+3

(c) y = 1/3 f(x)

(d) y = − f(x+4)

(e) y = 2f(x+6)

 

12. The graph of ƒ is given. Draw the graphs of the following functions.

(a) y = f(x+4)

(b) y = f(x) + 4

(c) y = 2f(x)

(d) y = − (½)f(x) + 3

 

13. The graph of ƒ is given. Use it to graph the following functions.

(a) y = f(2x)

(b) y = f( 1/2 x)

(c) y = f(‒x)

(d) y = ‒f(‒x)

 

14. The graph of y = √[3x‒x²] is given.

Use transformations to create a function whose graph is as shown.

 

15. (a) How is the graph of y = 2 sin x related to the graph of y = sinx? Sketch the graph y = 2 sin x.

(b) How is the graph of y = 1+ √x related to the graph of y = √x? Sketch the graph of y = 1 + √x.

 

16. Graph the function by hand, not by plotting points, but by starting with the graph of one of the standard functions given in Section 2, and then applying the appropriate transformations.

(a) y = ‒x³

(b) y = 1‒x²

(c) y = (x + 1)²

(d) y = x² ‒ 4x + 3

(e) y = 1 + 2cos x

(f) y = 4sin3x

(g) y = sin(x/2)

(h) y = 1 / x‒4

(i) y = √[x +3]

(j) y = (x + 2)⁴ + 3

(k) y = ½ (x² + 8x)

(l) y = 1 + ³√(x – 1)

(m) y = 2 / x+1

(n) y = ¼ tan (x − π/4)

(o) y = |sinx|

(p) y = |x² ‒ 2x|

 

17. The city of New Orleans is located at latitude 30°N. Use Figure 9 to find a function that models the number of hours of daylight at New Orleans as a function of the time of year. To check the accuracy of your model, use the fact that on March 31 the sun rises at 5:51 AM and sets at 6:18 PM in New Orleans.

 

18. A variable star is one whose brightness alternately increases and decreases. For the most visible variable star, Delta Cephei, the time between periods of maximum brightness is 54 days. the average brightness (or magnitude) of the star is 4.0, and its brightness varies by ±0.35 magnitude. Find a function that models the brightness of Delta Cephei as a function of time.

 

19. (a) How is the graph of y = f(|x|) related to the graph of f?

(b) Sketch the graph of y = sin |x|.

(c) Sketch the graph of y = √|x|.

 

20. Use the given graph of ƒ to sketch the graph of y = 1/f(x). Which features of f are the most important in sketching y = 1/ f(x) ? Explain how they are used.