Let g : R → R defined by the equation g(x) = x 2 + x. Let H ⊆ R and H = {y ∈ R : 6 ≤ y ≤ 12}. Then determine the inverse image g −1 (H).

Given g : R → R defined by the equation $g(x) = x^2 + x$ .

Given H ⊆ R and H = {y ∈ R : 6 ≤ y ≤ 12}.

As $y = x^2+x \implies y=x(x+1)$ .

Now, when $y = 6 \implies x(x+1) = 6 \implies x = -3 \ or \ x = 2$ .

Similarly when $y= 12 \implies x(x+1) = 12 \implies x = -4 \ or \ x = 3$ .

Also, $y' = 2x+1 > 0$ when $x > - \frac{1}{2}$ . Hence given function is increasing function for $x > - \frac{1}{2}$ .

And $y' < 0$ when $x < - \frac{1}{2}$ . Hence given function is decreasing function for $x < - \frac{1}{2}$ .

So, Set H under mapping $g^{-1}$, maps to $[-4,-3] \cup [2,3]$ .

Hence, $g^{-1}(H) = [-4,-3] \cup [2,3]$ .