Find expert answers and community support for all your questions on IDNLearn.com. Ask anything and get well-informed, reliable answers from our knowledgeable community members.
Sagot :
Sure! Let's verify the values of the function [tex]\( x^2 - 2x \)[/tex] at [tex]\( x = 2 \)[/tex] and [tex]\( x = 0 \)[/tex].
### Step-by-Step Solution
1. Define the Function:
[tex]\[ f(x) = x^2 - 2x \][/tex]
2. Calculate the Function Value at [tex]\( x = 2 \)[/tex]:
- Substitute [tex]\( x = 2 \)[/tex] into the function:
[tex]\[ f(2) = 2^2 - 2 \cdot 2 \][/tex]
- Perform the calculations:
[tex]\[ = 4 - 4 \][/tex]
- Simplify the result:
[tex]\[ = 0 \][/tex]
3. Calculate the Function Value at [tex]\( x = 0 \)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 0^2 - 2 \cdot 0 \][/tex]
- Perform the calculations:
[tex]\[ = 0 - 0 \][/tex]
- Simplify the result:
[tex]\[ = 0 \][/tex]
Therefore, the values of the function [tex]\( x^2 - 2x \)[/tex] at [tex]\( x = 2 \)[/tex] and [tex]\( x = 0 \)[/tex] are both:
[tex]\[ f(2) = 0 \][/tex]
[tex]\[ f(0) = 0 \][/tex]
So, the verified values are [tex]\( (0, 0) \)[/tex].
### Step-by-Step Solution
1. Define the Function:
[tex]\[ f(x) = x^2 - 2x \][/tex]
2. Calculate the Function Value at [tex]\( x = 2 \)[/tex]:
- Substitute [tex]\( x = 2 \)[/tex] into the function:
[tex]\[ f(2) = 2^2 - 2 \cdot 2 \][/tex]
- Perform the calculations:
[tex]\[ = 4 - 4 \][/tex]
- Simplify the result:
[tex]\[ = 0 \][/tex]
3. Calculate the Function Value at [tex]\( x = 0 \)[/tex]:
- Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 0^2 - 2 \cdot 0 \][/tex]
- Perform the calculations:
[tex]\[ = 0 - 0 \][/tex]
- Simplify the result:
[tex]\[ = 0 \][/tex]
Therefore, the values of the function [tex]\( x^2 - 2x \)[/tex] at [tex]\( x = 2 \)[/tex] and [tex]\( x = 0 \)[/tex] are both:
[tex]\[ f(2) = 0 \][/tex]
[tex]\[ f(0) = 0 \][/tex]
So, the verified values are [tex]\( (0, 0) \)[/tex].
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.