IDNLearn.com: Where your questions meet expert advice and community insights. Find the answers you need quickly and accurately with help from our knowledgeable and dedicated community members.
Sagot :
Let's solve the given problems step-by-step using algebraic methods.
### Problem 1:
We have two functions:
[tex]\[ f(x) = 2x + 3 \][/tex]
[tex]\[ g(x) = 3x - 2 \][/tex]
We need to find [tex]\((f + g)(2)\)[/tex] and [tex]\((f - g)(-2)\)[/tex].
#### (f + g)(2):
First, evaluate [tex]\(f(2)\)[/tex]:
[tex]\[ f(2) = 2(2) + 3 = 4 + 3 = 7 \][/tex]
Next, evaluate [tex]\(g(2)\)[/tex]:
[tex]\[ g(2) = 3(2) - 2 = 6 - 2 = 4 \][/tex]
Now, add the results:
[tex]\[ (f + g)(2) = f(2) + g(2) = 7 + 4 = 11 \][/tex]
#### (f - g)(-2):
First, evaluate [tex]\(f(-2)\)[/tex]:
[tex]\[ f(-2) = 2(-2) + 3 = -4 + 3 = -1 \][/tex]
Next, evaluate [tex]\(g(-2)\)[/tex]:
[tex]\[ g(-2) = 3(-2) - 2 = -6 - 2 = -8 \][/tex]
Now, subtract the results:
[tex]\[ (f - g)(-2) = f(-2) - g(-2) = -1 - (-8) = -1 + 8 = 7 \][/tex]
### Problem 2:
We have two functions:
[tex]\[ f(x) = x^2 - 3x + 2 \][/tex]
[tex]\[ g(x) = 2x^2 - x + 1 \][/tex]
We need to find [tex]\((f + g)(2)\)[/tex] and [tex]\((f - g)(-2)\)[/tex].
#### (f + g)(2):
First, evaluate [tex]\(f(2)\)[/tex]:
[tex]\[ f(2) = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0 \][/tex]
Next, evaluate [tex]\(g(2)\)[/tex]:
[tex]\[ g(2) = 2(2^2) - 2 + 1 = 2 \cdot 4 - 2 + 1 = 8 - 2 + 1 = 7 \][/tex]
Now, add the results:
[tex]\[ (f + g)(2) = f(2) + g(2) = 0 + 7 = 7 \][/tex]
#### (f - g)(-2):
First, evaluate [tex]\(f(-2)\)[/tex]:
[tex]\[ f(-2) = (-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12 \][/tex]
Next, evaluate [tex]\(g(-2)\)[/tex]:
[tex]\[ g(-2) = 2(-2)^2 - (-2) + 1 = 2 \cdot 4 + 2 + 1 = 8 + 2 + 1 = 11 \][/tex]
Now, subtract the results:
[tex]\[ (f - g)(-2) = f(-2) - g(-2) = 12 - 11 = 1 \][/tex]
### Summary:
- For [tex]\(f(x) = 2x + 3\)[/tex] and [tex]\(g(x) = 3x - 2\)[/tex]:
[tex]\[ (f + g)(2) = 11, \quad (f - g)(-2) = 7 \][/tex]
- For [tex]\(f(x) = x^2 - 3x + 2\)[/tex] and [tex]\(g(x) = 2x^2 - x + 1\)[/tex]:
[tex]\[ (f + g)(2) = 7, \quad (f - g)(-2) = 1 \][/tex]
### Problem 1:
We have two functions:
[tex]\[ f(x) = 2x + 3 \][/tex]
[tex]\[ g(x) = 3x - 2 \][/tex]
We need to find [tex]\((f + g)(2)\)[/tex] and [tex]\((f - g)(-2)\)[/tex].
#### (f + g)(2):
First, evaluate [tex]\(f(2)\)[/tex]:
[tex]\[ f(2) = 2(2) + 3 = 4 + 3 = 7 \][/tex]
Next, evaluate [tex]\(g(2)\)[/tex]:
[tex]\[ g(2) = 3(2) - 2 = 6 - 2 = 4 \][/tex]
Now, add the results:
[tex]\[ (f + g)(2) = f(2) + g(2) = 7 + 4 = 11 \][/tex]
#### (f - g)(-2):
First, evaluate [tex]\(f(-2)\)[/tex]:
[tex]\[ f(-2) = 2(-2) + 3 = -4 + 3 = -1 \][/tex]
Next, evaluate [tex]\(g(-2)\)[/tex]:
[tex]\[ g(-2) = 3(-2) - 2 = -6 - 2 = -8 \][/tex]
Now, subtract the results:
[tex]\[ (f - g)(-2) = f(-2) - g(-2) = -1 - (-8) = -1 + 8 = 7 \][/tex]
### Problem 2:
We have two functions:
[tex]\[ f(x) = x^2 - 3x + 2 \][/tex]
[tex]\[ g(x) = 2x^2 - x + 1 \][/tex]
We need to find [tex]\((f + g)(2)\)[/tex] and [tex]\((f - g)(-2)\)[/tex].
#### (f + g)(2):
First, evaluate [tex]\(f(2)\)[/tex]:
[tex]\[ f(2) = 2^2 - 3(2) + 2 = 4 - 6 + 2 = 0 \][/tex]
Next, evaluate [tex]\(g(2)\)[/tex]:
[tex]\[ g(2) = 2(2^2) - 2 + 1 = 2 \cdot 4 - 2 + 1 = 8 - 2 + 1 = 7 \][/tex]
Now, add the results:
[tex]\[ (f + g)(2) = f(2) + g(2) = 0 + 7 = 7 \][/tex]
#### (f - g)(-2):
First, evaluate [tex]\(f(-2)\)[/tex]:
[tex]\[ f(-2) = (-2)^2 - 3(-2) + 2 = 4 + 6 + 2 = 12 \][/tex]
Next, evaluate [tex]\(g(-2)\)[/tex]:
[tex]\[ g(-2) = 2(-2)^2 - (-2) + 1 = 2 \cdot 4 + 2 + 1 = 8 + 2 + 1 = 11 \][/tex]
Now, subtract the results:
[tex]\[ (f - g)(-2) = f(-2) - g(-2) = 12 - 11 = 1 \][/tex]
### Summary:
- For [tex]\(f(x) = 2x + 3\)[/tex] and [tex]\(g(x) = 3x - 2\)[/tex]:
[tex]\[ (f + g)(2) = 11, \quad (f - g)(-2) = 7 \][/tex]
- For [tex]\(f(x) = x^2 - 3x + 2\)[/tex] and [tex]\(g(x) = 2x^2 - x + 1\)[/tex]:
[tex]\[ (f + g)(2) = 7, \quad (f - g)(-2) = 1 \][/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.