Functions Exercise: Finding Function Values and Function Compositions, Cheat Sheet of Calculus

Solutions to exercise 1.1 of a functions course, which involves finding function values and function compositions for various functions using the given formulas.

Typology: Cheat Sheet

2020/2021

Uploaded on 10/25/2021

Erwin_Rommel
Erwin_Rommel ๐Ÿ‡ฌ๐Ÿ‡ง

5

(1)

3 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Name:
Score:
Course/Yr.&Sec.:
Date:
EXERCISE 1.1 FUNCTIONS
A. For each of the following functions,
1. f(x) = 7x
2
+ 4x โ€“ 1
a) f (-1/3) = 7(-1/3)
2
+ 4(-1/3) โ€“ 1
f (-1/3) = 7/9- 4/3 โ€“ 1 = 7/9 โ€“ 12/9 -9/9 = -14/9
b) f (5) = 7(5)
2
+ 4(5) โ€“ 1
f (5) = 175 + 20 โ€“ 1 = 194
c) f(7x
2
) = 7(7x
2
)
2
+ 4(7x
2
) โ€“ 1
f(7x
2
) = 7(49x
4
) + 28x
2
โ€“ 1 = 343x
4
+ 28x
2
-1
d) f(x) + f(h) = 7(x)
2
+ 4(x) โ€“ 1 + [7(h)
2
+ 4(h) โ€“ 1]
f(x) + f(h) = 7x
2
+ 4x โ€“ 1 + 7h
2
+ 4h - 1
f(x) + f(h) = 7x
2
+ 4x + 7h
2
+ 4h -2
f(x) + f(h) = 7h
2
+ 7x
2
+4h+4x -2
e) f(x
2
) +f(h
2
) = 7(x
2
)
2
+ 4(x
2
) โ€“ 1 โ€“[7(h
2
)
2
+ 4(h
2
)-1]
f(x
2
) +f(h
2
) = 7x
4
+ 4x
2
โ€“ 1 โ€“ (7 h
4
+ 4h
2
โ€“ 1)
f(x
2
) +f(h
2
) = 7x
4
+ 4x
2
โ€“ 1 โ€“ 7h
4
โ€“ 4h
2
+1
f(x
2
) +f(h
2
) = 7x
4
+ 4x
2
โ€“ 7h
4
- 4h
2
f(x
2
) +f(h
2
) = 7x
4
โ€“ 7h
4
- 4h
2
+ 4x
2
2. f(x) = 7x
3
- 5
a) f (-1/3) = 7(-1/3)
3
-5
f (-1/3) = 7(-1/27) -5 = -142/27
b) f (5) = 7(5)
3
-5
f (5) = 7(125) -5 = 870
c) f(7x
2
) = 7(7x
2
)
3
-5
f(7x
2
) = 7(343x
6
) -5 = 2401 x
6
-5
d) f(x) + f(h) = 7(x)
3
-5 + [7(h)
3
-5]
f(x) + f(h) = 7x
3
- 5 + 7h
3
-5
f(x) + f(h) = 7x
3
-5 +7h
3
- 5
f(x) + f(h) = 7x
3
+ 7h
3
-10
f(x) + f(h) = 7(x
3
+ h
3
) -10
e) f(x
2
) - f(h
2
) = 7(x
2
)
3
โ€“ 5 โ€“ [7(h
2
)
3
-5]
f(x
2
) - f(h
2
) = 7x
6
โ€“ 5 โ€“ (7h
6
-5)
f(x
2
) - f(h
2
) = 7x
6
โ€“ 5 โ€“ 7h
6
+ 5
pf3
pf4

Partial preview of the text

Download Functions Exercise: Finding Function Values and Function Compositions and more Cheat Sheet Calculus in PDF only on Docsity!

Name: Course/Yr.&Sec.: Score:Date:

EXERCISE 1.1 FUNCTIONS A. For each of the following functions,

  1. f(x) = 7x^2 + 4x โ€“ 1 a) f (-1/3) = 7(-1/3)^2 + 4(-1/3) โ€“ 1 f (-1/3) = 7/9- 4/3 โ€“ 1 = 7/9 โ€“ 12/9 -9/9 = -14/ b) f (5) = 7(5)f (5) = 175 + 20 โ€“ 1 = 194^2 + 4(5) โ€“ 1 c) f(7x^2 ) = 7(7x^2 )^2 + 4(7x^2 ) โ€“ 1 f(7x^2 ) = 7(49x^4 ) + 28x^2 โ€“ 1 = 343x^4 + 28x^2 - d) f(x) + f(h) = 7(x)f(x) + f(h) = 7x (^2) + 4x โ€“ 1 + 7h^2 + 4(x) โ€“ 1 + [7(h) (^2) + 4h - 1^2 + 4(h) โ€“ 1] f(x) + f(h) = 7xf(x) + f(h) = 7h^22 + 4x + 7h+ 7x (^2) +4h+4x -2^2 + 4h - e) f(x^2 ) +f(h^2 ) = 7(x^2 )^2 + 4(x^2 ) โ€“ 1 โ€“[7(h^2 )^2 + 4(h^2 )-1] f(xf(x^22 ) +f(h) +f(h^22 ) = 7x) = 7x^44 + 4x+ 4x^22 โ€“ 1 โ€“ (7 hโ€“ 1 โ€“ 7h (^4 4) โ€“ 4h+ 4h (^22) +1โ€“ 1) f(x f(x^22 ) +f(h) +f(h^22 ) = 7x) = 7x^44 + 4xโ€“ 7h^24 โ€“ 7h- 4h 24 - 4h+ 4x^22
  2. f(x) = 7x^3 - 5 a) f (-1/3) = 7(-1/3)f (-1/3) = 7(-1/27) -5 = -142/27^3 -

b) f (5) = 7(5)f (5) = 7(125) -5 = 870^3 - c) f(7x f(7x^22 ) = 7(7x) = 7(343x^2 )^3 6 -5) -5 = 2401 x (^6) - d) f(x) + f(h) = 7(x)^3 -5 + [7(h)^3 -5] f(x) + f(h) = 7xf(x) + f(h) = 7x^33 - 5 + 7h-5 +7h (^3 3) - 5^ - f(x) + f(h) = 7xf(x) + f(h) = 7(x^3 3 + 7h (^) + h (^33) ) -10- e) f(x^2 ) - f(h^2 ) = 7(x^2 )^3 โ€“ 5 โ€“ [7(h^2 )^3 -5] f(xf(x^22 ) - f(h) - f(h^22 ) = 7x) = 7x^66 โ€“ 5 โ€“ (7hโ€“ 5 โ€“ 7h (^66) + 5-5)

f(x^2 ) - f(h^2 ) = 7x^6 โ€“ 7h^6 f(x^2 ) - f(h^2 ) = 7(x^6 โ€“ h^6 ) Find: a. f (-1/3) b. f (5) c. f(7x^2 ) d. f(x) + f(h) e. f(x^2 ) โ€“ f(h^2 )

B. Find and simplify  for each of the following functions,

  1. g(x) = x + 7

โ„Ž =^

= 7 โˆ’  + โ„Žโ„Ž^ โˆ’ โˆš7 โˆ’  = โˆš7 โˆ’  โˆ’ โ„Žโ„Ž^ โˆ’ โˆš7 โˆ’  โˆ™ โˆš7 โˆ’  โˆ’ โ„Žโˆš7 โˆ’  โˆ’ โ„Ž^ + โˆš7 โˆ’ + โˆš7 โˆ’  = (^) โ„Ž โˆš7 โˆ’  โˆ’ โ„Ž7 โˆ’  โˆ’ โ„Ž โˆ’ 7 โˆ’  + โˆš7 โˆ’  = (^) โ„Ž โˆš7 โˆ’  โˆ’ โ„Žโˆ’โ„Ž + โˆš7 โˆ’ 

= (^) โˆš 7 โˆ’  โˆ’ โ„Žโˆ’1 + โˆš 7 โˆ’  = (^) โˆš7 โˆ’  + โˆš7 โˆ’ โˆ’1  = (^) 2โˆš7 โˆ’ โˆ’

  1. g(x) = 5x^2 -2x+

5 + โ„Ž^ โˆ’ 2 + โ„Ž + 3 โˆ’ 5^ โˆ’ 2 + 3

5^ + 2โ„Ž + โ„Ž โˆ’ 2 โˆ’ 2โ„Ž + 3 โˆ’ 5^ โˆ’ 2 + 3

= 5^ + 10โ„Ž + 5โ„Ž^ โˆ’ 2^ โ„Žโˆ’ 2โ„Ž^ + 3 โˆ’ 5^ + 2 โˆ’ 3

10โ„Ž + 5โ„Ž^ โˆ’ 2โ„Ž

  1. f o g (2)

internal function g(x) = 4x โ€“ 3 (^) ๏ƒ  g(2) = 4(2) โ€“ 3 = 5 external function f(x)= 2x^2 -3x + 4 f o g (2) = f[g (2)] = 2(5)^2 -3(5) + 4 = 39

  1. f [g (2)]

internal function g(x) = 4x โ€“ 3 (^) ๏ƒ  g(2) = 4(2) โ€“ 3 = 5 external function f(x)= 2x^2 -3x + 4 f o g (2) = f[g (2)] = 2(5)^2 -3(5) + 4 = 39

  1. f o g (-1)

internal function g(x) = 4x โ€“ 3 ๏ƒ  g(-1) = 4(-1) โ€“ 3 = - external function f(x)= 2x^2 -3x + 4 f o g (2) = f[g (2)] = 2(-7)^2 -3(-7) + 4 = 123

  1. g o g (-2) internal function g(x) = 4x โ€“ 3 ๏ƒ  g(-2) = 4(-2) โ€“ 3 = - external function g(x) = 4x โ€“ 3 g o g (-2) = g[g (-2)] = 4(-11) โ€“ 3 = -