Question

score: 0/5 answered: 0/5
question 1
let
$f(x)=\frac{1}{x + 11}$
$f^{-1}(x)=
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question 2
let $f(x)=3+sqrt{3x - 2}$. find $f^{-1}(x)$.
$f^{-1}(x)=quad$, for $xgeq3$
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question 3
find the inverse function of $f(x)=10+sqrt3{x}$.
$f^{-1}(x)=
$

Explanation:

Question 1

Step1: Replace $f(x)$ with $y$

$y = \frac{1}{x + 11}$

Step2: Swap $x$ and $y$

$x=\frac{1}{y + 11}$

Step3: Solve for $y$

$x(y + 11)=1$; $xy+11x = 1$; $xy=1 - 11x$; $y=\frac{1 - 11x}{x}=\frac{1}{x}-11$ ($x
eq0$)

Step1: Replace $f(x)$ with $y$

$y = 3+\sqrt{3x - 2}$

Step2: Swap $x$ and $y$

$x = 3+\sqrt{3y - 2}$

Step3: Isolate the square - root term

$x - 3=\sqrt{3y - 2}$

Step4: Square both sides

$(x - 3)^2=3y - 2$

Step5: Solve for $y$

$3y=(x - 3)^2 + 2$; $y=\frac{(x - 3)^2+2}{3}=\frac{x^{2}-6x + 9+2}{3}=\frac{x^{2}-6x + 11}{3}$

Step1: Replace $f(x)$ with $y$

$y = 10+\sqrt[3]{x}$

Step2: Swap $x$ and $y$

$x = 10+\sqrt[3]{y}$

Step3: Isolate the cube - root term

$x - 10=\sqrt[3]{y}$

Step4: Cube both sides

$(x - 10)^3=y$

Answer:

$f^{-1}(x)=\frac{1 - 11x}{x},x
eq0$

Question 2