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MTH100 ASSIGNMENT NO. 1 FALL 2022 |
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MTH100 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || GENERAL MATHEMATICS || BY VuTech
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Fall
2022 MTH100:
Assignment
No. 1 Total Marks: 10
Due
Date: 01-12- 2022
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Question
No.1 Marks: 10
If `\f\left(x\right) = \sqrt{x - 3}` then verify
`\left(fof^{-1}\right)\left(x\right) = \left(f^- 1of\right)\left(x\right) =
x`
Answer:
Here `f\left(x \right) = \sqrt {x - 3}`
Let assume `f\left(x\right) = y`
then `y = \sqrt {x - 3}`
`y^2 = x - 3`
`x = y^2 + 3`
We know that `\(f^ - 1\left(y\right) = x`
`\(f^- 1\left(y\right) = y^2 + 3`
By replacing y by x then `\(f^- 1\left(x\right) =
x^2 + 3`
First we solve `\left(fof^-
1\right)\left(x\right)`
`\left(fof^- 1\right)\left(x \right) = f\left(f^ -
1\left(x\right)\right)`
`\left(fof^- 1\right)\left( x \right) = f\left(x^2
+ 3\right)`
`\left(fof^- 1\right)\left(x\right) =
\sqrt{\left(x^2 + 3\right)- 3}`
`\left(fof^- 1\right)\left(x\right) = \sqrt
x^2`
`\left(fof^ - 1\right)\left(x\right) = x
- - - - - - - \left(1\right)`
Now we solve `\left(f^-
1of\right)\left(x\right)`
`\left(f^- 1of\right)\left(x\right) = f^-
1\left(f\left(x\right)\right)`
`\left(f^- 1of\right)\left(x\right) = f^-
1\left(\sqrt {x - 3}\right)`
`\left(f^- 1of\right)\left(x\right) = \left(\sqrt{x
- 3}\right)^2 + 3`
`\left(f^- 1of\right)\left( x \right) = x - 3 +
3`
`\left(f^- 1of\right)\left(x\right) = x
- - - - - - - - - -\left( 2
\right)`
By (1) and (2) we can prove that
`\left(fof^- 1\right)\left(x\right) = \left(f^- 1of
\right)\left(x\right) = x`
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