Find the Inverse f(x)=1+ square root of 2+3x

Problem

ƒ(x)=1+√(,2+3*x)

Solution

1. Replace ƒ(x) with y to make the algebraic manipulation easier.

y=1+√(,2+3*x)

2. Swap the variables x and y to begin solving for the inverse function.

x=1+√(,2+3*y)

3. Isolate the radical term by subtracting 1 from both sides of the equation.

x−1=√(,2+3*y)

4. Square both sides of the equation to eliminate the square root.

(x−1)2=2+3*y

5. Isolate the term containing y by subtracting 2 from both sides.

(x−1)2−2=3*y

6. Solve for y by dividing the entire expression by 3

y=((x−1)2−2)/3

7. Define the domain of the inverse function based on the range of the original function. Since √(,2+3*x)≥0 then ƒ(x)≥1

x≥1

Final Answer

ƒ(x)(−1)=((x−1)2−2)/3,x≥1

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