Theory of Intelligence
11TH•Mathematics•Theory Of Quadratic Functions•Question 3

Step-by-Step Solution

Find the inverse of the following quadratic functions. Also find their domain and range:

Step-by-Step Solution

Step 1: Analyze Behavior on the Domain & Determine Range

The vertex of the full quadratic occurs at:

h=−b2a=−−22(3)=13h = -\frac{b}{2a} = -\frac{-2}{2(3)} = \frac{1}{3}

Because the restricted domain is x≥5x \ge 5, which lies entirely to the right of the vertex x=13x = \frac{1}{3}, the function f(x)f(x) is strictly increasing on [5,∞)[5, \infty).

The minimum value on this interval occurs at the left boundary x=5x = 5:

f(5)=3(5)2−2(5)+6=3(25)−10+6=75−4=71f(5) = 3(5)^2 - 2(5) + 6 = 3(25) - 10 + 6 = 75 - 4 = 71

As x→∞x \to \infty, f(x)→∞f(x) \to \infty. Therefore, the range is:

Ran(f)=[71,∞)\text{Ran}(f) = [71, \infty)

Step 2: Solve for xx in Terms of yy

Set y=3x2−2x+6y = 3x^2 - 2x + 6, yielding the quadratic equation in xx:

3x2−2x+(6−y)=03x^2 - 2x + (6 - y) = 0

Applying the quadratic formula with A=3A = 3, B=−2B = -2, C=6−yC = 6 - y:

x=−(−2)±(−2)2−4(3)(6−y)2(3)=2±4−72+12y6=2±12y−686=2±4(3y−17)6=2±23y−176=1±3y−173\begin{aligned} x &= \frac{-(-2) \pm\sqrt{(-2)^2 - 4(3)(6 - y)}}{2(3)} \\ &= \frac{2 \pm\sqrt{4 - 72 + 12y}}{6} \\ &= \frac{2 \pm\sqrt{12y - 68}}{6} \\ &= \frac{2 \pm\sqrt{4(3y - 17)}}{6} \\ &= \frac{2 \pm 2\sqrt{3y - 17}}{6} \\ &= \frac{1 \pm\sqrt{3y - 17}}{3} \end{aligned}

Step 3: Justify the Selection of the Radical Sign (±\pm)

At the boundary value y=71y = 71:

  • For the positive sign: x=1+3(71)−173=1+213−173=1+1963=1+143=153=5x = \frac{1 + \sqrt{3(71) - 17}}{3} = \frac{1 + \sqrt{213 - 17}}{3} = \frac{1 + \sqrt{196}}{3} = \frac{1 + 14}{3} = \frac{15}{3} = 5. This satisfies x≥5x \ge 5.
  • For the negative sign: x=1−143=−133≈−4.33<5x = \frac{1 - 14}{3} = -\frac{13}{3} \approx -4.33 < 5, violating the domain condition.

Hence, we must select the positive root:

x=1+3y−173x = \frac{1 + \sqrt{3y - 17}}{3}

Step 4: Swap Variables and State Domains

Interchanging variables:

f−1(x)=1+3x−173f^{-1}(x) = \frac{1 + \sqrt{3x - 17}}{3}

The swapped intervals are:

Dom(f−1)=Ran(f)=[71,∞)Ran(f−1)=Dom(f)=[5,∞)\begin{aligned} \text{Dom}(f^{-1}) &= \text{Ran}(f) = [71, \infty) \\ \text{Ran}(f^{-1}) &= \text{Dom}(f) = [5, \infty) \end{aligned}

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