Step-by-Step Solution
Step 1: Analyze Behavior on the Domain & Determine Range
The vertex of the full quadratic occurs at:
h=−2ab=−2(3)−2=31
Because the restricted domain is x≥5, which lies entirely to the right of the vertex x=31, the function f(x) is strictly increasing on [5,∞).
The minimum value on this interval occurs at the left boundary x=5:
f(5)=3(5)2−2(5)+6=3(25)−10+6=75−4=71
As x→∞, f(x)→∞. Therefore, the range is:
Ran(f)=[71,∞)
Step 2: Solve for x in Terms of y
Set y=3x2−2x+6, yielding the quadratic equation in x:
3x2−2x+(6−y)=0
Applying the quadratic formula with A=3, B=−2, C=6−y:
x=2(3)−(−2)±(−2)2−4(3)(6−y)=62±4−72+12y=62±12y−68=62±4(3y−17)=62±23y−17=31±3y−17
Step 3: Justify the Selection of the Radical Sign (±)
At the boundary value y=71:
- For the positive sign: x=31+3(71)−17=31+213−17=31+196=31+14=315=5. This satisfies x≥5.
- For the negative sign: x=31−14=−313≈−4.33<5, violating the domain condition.
Hence, we must select the positive root:
x=31+3y−17
Step 4: Swap Variables and State Domains
Interchanging variables:
f−1(x)=31+3x−17
The swapped intervals are:
Dom(f−1)Ran(f−1)=Ran(f)=[71,∞)=Dom(f)=[5,∞)