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Sure, let's solve each part step-by-step given the function [tex]\( f(x) = x^2 - 3x + 4 \)[/tex].
1. Find [tex]\( f(0) \)[/tex]:
To find [tex]\( f(0) \)[/tex], we substitute [tex]\( x = 0 \)[/tex] into the function [tex]\( f(x) \)[/tex].
[tex]\[ f(0) = (0)^2 - 3(0) + 4 = 4 \][/tex]
Therefore, [tex]\( f(0) = 4 \)[/tex].
2. Find [tex]\( f(2x) \)[/tex]:
To find [tex]\( f(2x) \)[/tex] for [tex]\( x = 2 \)[/tex]:
First, consider [tex]\( f(2x) \)[/tex]:
[tex]\[ f(2x) = (2x)^2 - 3(2x) + 4 \][/tex]
Substituting [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2 \cdot 2) = (2 \cdot 2)^2 - 3(2 \cdot 2) + 4 \][/tex]
[tex]\[ f(4) = 4^2 - 3 \cdot 4 + 4 \][/tex]
[tex]\[ f(4) = 16 - 12 + 4 \][/tex]
[tex]\[ f(4) = 8 \][/tex]
Therefore, [tex]\( f(2x) \)[/tex] when [tex]\( x = 2 \)[/tex] is 8.
3. Find [tex]\( f(-1) \)[/tex]:
To find [tex]\( f(-1) \)[/tex], we substitute [tex]\( x = -1 \)[/tex] into the function [tex]\( f(x) \)[/tex].
[tex]\[ f(-1) = (-1)^2 - 3(-1) + 4 \][/tex]
[tex]\[ f(-1) = 1 + 3 + 4 \][/tex]
[tex]\[ f(-1) = 8 \][/tex]
Therefore, [tex]\( f(-1) = 8 \)[/tex].
4. Find [tex]\( f(-3) \)[/tex]:
To find [tex]\( f(-3) \)[/tex], we substitute [tex]\( x = -3 \)[/tex] into the function [tex]\( f(x) \)[/tex].
[tex]\[ f(-3) = (-3)^2 - 3(-3) + 4 \][/tex]
[tex]\[ f(-3) = 9 + 9 + 4 \][/tex]
[tex]\[ f(-3) = 22 \][/tex]
Therefore, [tex]\( f(-3) = 22 \)[/tex].
5. Find [tex]\( f(y+1) \)[/tex]:
To find [tex]\( f(y+1) \)[/tex] for [tex]\( y = 1 \)[/tex]:
First, consider [tex]\( f(y+1) \)[/tex]:
[tex]\[ f(y+1) = (y+1)^2 - 3(y+1) + 4 \][/tex]
Substituting [tex]\( y = 1 \)[/tex]:
[tex]\[ f(1+1) = (1+1)^2 - 3(1+1) + 4 \][/tex]
[tex]\[ f(2) = 2^2 - 3 \cdot 2 + 4 \][/tex]
[tex]\[ f(2) = 4 - 6 + 4 \][/tex]
[tex]\[ f(2) = 2 \][/tex]
Therefore, [tex]\( f(y+1) \)[/tex] when [tex]\( y = 1 \)[/tex] is 2.
6. Find [tex]\( f\left(\frac{1}{x}\right) \)[/tex]:
To find [tex]\( f\left(\frac{1}{x}\right) \)[/tex] for [tex]\( x = 1 \)[/tex]:
First, consider [tex]\( f\left(\frac{1}{x}\right) \)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 - 3 \left(\frac{1}{x}\right) + 4 \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ f\left(\frac{1}{1}\right) = \left(\frac{1}{1}\right)^2 - 3 \left(\frac{1}{1}\right) + 4 \][/tex]
[tex]\[ f(1) = 1^2 - 3 \cdot 1 + 4 \][/tex]
[tex]\[ f(1) = 1 - 3 + 4 \][/tex]
[tex]\[ f(1) = 2 \][/tex]
Therefore, [tex]\( f\left(\frac{1}{x}\right) \)[/tex] when [tex]\( x = 1 \)[/tex] is 2.0 (which is mathematically the same as 2).
Hence, the values are:
1. [tex]\( f(0) = 4 \)[/tex]
2. [tex]\( f(2x) = 8 \)[/tex] when [tex]\( x = 2 \)[/tex]
3. [tex]\( f(-1) = 8 \)[/tex]
4. [tex]\( f(-3) = 22 \)[/tex]
5. [tex]\( f(y+1) = 2 \)[/tex] when [tex]\( y = 1 \)[/tex]
6. [tex]\( f\left(\frac{1}{x}\right) = 2.0 \)[/tex] when [tex]\( x = 1 \)[/tex]
1. Find [tex]\( f(0) \)[/tex]:
To find [tex]\( f(0) \)[/tex], we substitute [tex]\( x = 0 \)[/tex] into the function [tex]\( f(x) \)[/tex].
[tex]\[ f(0) = (0)^2 - 3(0) + 4 = 4 \][/tex]
Therefore, [tex]\( f(0) = 4 \)[/tex].
2. Find [tex]\( f(2x) \)[/tex]:
To find [tex]\( f(2x) \)[/tex] for [tex]\( x = 2 \)[/tex]:
First, consider [tex]\( f(2x) \)[/tex]:
[tex]\[ f(2x) = (2x)^2 - 3(2x) + 4 \][/tex]
Substituting [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2 \cdot 2) = (2 \cdot 2)^2 - 3(2 \cdot 2) + 4 \][/tex]
[tex]\[ f(4) = 4^2 - 3 \cdot 4 + 4 \][/tex]
[tex]\[ f(4) = 16 - 12 + 4 \][/tex]
[tex]\[ f(4) = 8 \][/tex]
Therefore, [tex]\( f(2x) \)[/tex] when [tex]\( x = 2 \)[/tex] is 8.
3. Find [tex]\( f(-1) \)[/tex]:
To find [tex]\( f(-1) \)[/tex], we substitute [tex]\( x = -1 \)[/tex] into the function [tex]\( f(x) \)[/tex].
[tex]\[ f(-1) = (-1)^2 - 3(-1) + 4 \][/tex]
[tex]\[ f(-1) = 1 + 3 + 4 \][/tex]
[tex]\[ f(-1) = 8 \][/tex]
Therefore, [tex]\( f(-1) = 8 \)[/tex].
4. Find [tex]\( f(-3) \)[/tex]:
To find [tex]\( f(-3) \)[/tex], we substitute [tex]\( x = -3 \)[/tex] into the function [tex]\( f(x) \)[/tex].
[tex]\[ f(-3) = (-3)^2 - 3(-3) + 4 \][/tex]
[tex]\[ f(-3) = 9 + 9 + 4 \][/tex]
[tex]\[ f(-3) = 22 \][/tex]
Therefore, [tex]\( f(-3) = 22 \)[/tex].
5. Find [tex]\( f(y+1) \)[/tex]:
To find [tex]\( f(y+1) \)[/tex] for [tex]\( y = 1 \)[/tex]:
First, consider [tex]\( f(y+1) \)[/tex]:
[tex]\[ f(y+1) = (y+1)^2 - 3(y+1) + 4 \][/tex]
Substituting [tex]\( y = 1 \)[/tex]:
[tex]\[ f(1+1) = (1+1)^2 - 3(1+1) + 4 \][/tex]
[tex]\[ f(2) = 2^2 - 3 \cdot 2 + 4 \][/tex]
[tex]\[ f(2) = 4 - 6 + 4 \][/tex]
[tex]\[ f(2) = 2 \][/tex]
Therefore, [tex]\( f(y+1) \)[/tex] when [tex]\( y = 1 \)[/tex] is 2.
6. Find [tex]\( f\left(\frac{1}{x}\right) \)[/tex]:
To find [tex]\( f\left(\frac{1}{x}\right) \)[/tex] for [tex]\( x = 1 \)[/tex]:
First, consider [tex]\( f\left(\frac{1}{x}\right) \)[/tex]:
[tex]\[ f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 - 3 \left(\frac{1}{x}\right) + 4 \][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ f\left(\frac{1}{1}\right) = \left(\frac{1}{1}\right)^2 - 3 \left(\frac{1}{1}\right) + 4 \][/tex]
[tex]\[ f(1) = 1^2 - 3 \cdot 1 + 4 \][/tex]
[tex]\[ f(1) = 1 - 3 + 4 \][/tex]
[tex]\[ f(1) = 2 \][/tex]
Therefore, [tex]\( f\left(\frac{1}{x}\right) \)[/tex] when [tex]\( x = 1 \)[/tex] is 2.0 (which is mathematically the same as 2).
Hence, the values are:
1. [tex]\( f(0) = 4 \)[/tex]
2. [tex]\( f(2x) = 8 \)[/tex] when [tex]\( x = 2 \)[/tex]
3. [tex]\( f(-1) = 8 \)[/tex]
4. [tex]\( f(-3) = 22 \)[/tex]
5. [tex]\( f(y+1) = 2 \)[/tex] when [tex]\( y = 1 \)[/tex]
6. [tex]\( f\left(\frac{1}{x}\right) = 2.0 \)[/tex] when [tex]\( x = 1 \)[/tex]
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