Answer:
None of the following statements have the same result.
Step-by-step explanation:
1) f(2) when f(x) = 3x + 4
We just have to put x=2
[tex]f(x)=3x+4\\f(2)=3(2)+4\\f(2)=6+4\\f(2)=10[/tex]
So, f(2) when f(x) = 3x + 4 is x=10
2) f⁻¹ (4) when f(x) =3x-4/5
We need to find f⁻¹(x) first.
Put [tex]y=3x-\frac{4}{5}[/tex]
Now solve for x
Add 4/5 on both sides
[tex]y+\frac{4}{5}=3x-\frac{4}{5}\\y+\frac{4}{5}=3x\\x=\frac{1}{3}y+\frac{4}{5*3}\\x=\frac{1}{3}y+\frac{4}{15}\\x=\frac{5y+4}{15}[/tex]
Now put f⁻¹(x) instead of x and replace y with x
[tex]f^{-1}(x)=\frac{5x+4}{15}[/tex]
Now finding f⁻¹(4)
[tex]f^{-1}(x)=\frac{5x+4}{15} \\f^{-1}(4)=\frac{5(4)+4}{15} \\f^{-1}(4)=\frac{20+4}{15} \\f^{-1}(4)=\frac{24}{15}[/tex]
f⁻¹ (4) when f(x) =3x-4/5 is 24/15
3) [tex]3y - 6 = y + 10\\[/tex]
Solving:
[tex]3y - 6 = y + 10\\3y-y=10+6\\2y=16\\y=16/2\\y=8[/tex]
Solving 3y − 6 = y + 10, we get y=8
None of the following statements have the same result.
