[size=150]Consider the function f : R → R defined as f (x) = x[sup]2[/sup] + 3. [br][br]Find f [−3,5] and f [sup]-1[/sup] [12,19].[br][br]Answers: f [−3,5] = [12,28]; [br] f [sup]-1[/sup] [12,19] = [−4,−3]∪[3,4].[/size]

T[size=150]o find the image of a set under a function and the pre-image of an interval under the inverse of a function, we need to apply the function and its inverse to the given set or interval, respectively.[br][br]Given the function [i]f: [/i]Z→Z defined as f(x) = x[sup]2[/sup] + 3, let's find the following:[br][br]a)Image of [-3, 5] under [i]f[/i], denoted as f[-3, 5]:[br]To find the image of the interval [-3, 5] under the function [i]f[/i], we need to apply f(x) to each value in the interval and collect the resulting values.[br]f(-3) = (-3)[sup]2[/sup] + 3 = 9 + 3 = 12[br]f(5) = 5[sup]2[/sup] + 3 = 25 + 3 = 28[br]The mage of [-3, 5] under [i]f[/i] is the interval [12, 28], denoted as f[-3, 5] = [12, 28].[br][br]b)Pre-image of [12, 19] under [i]f[/i][sup]--1[/sup],denoted as [i]f[sup]--1[/sup][/i][12, 19]:[br]To find the pre-image of the interval [12, 19] under the inverse function [i]f[/i][sup]-1[/sup], [br]we need to apply [i]f[/i][sup]--1[/sup](x)[br]to each value in the interval and collect the resulting values. For x in [12, 19], [br]we want to find [i]f[/i][sup]--1[/sup](x).[br]Given that f(x) = x[sup]2[/sup] + 3, we set x = x[sup]2[/sup] + 3 and solve for x[sup]2[/sup]:  x - 3 = x[sup]2[/sup][br]Now, we can use the quadratic formula to find x[sup]2[/sup]:[br]x[sup]2 = [math]\frac{-1\pm\sqrt{1+4x}}{2}[/math][/sup][br][br]Since we are only interested in the real values of x, we consider the positive and[br]negative solutions separately:[br]  For x: [img width=10,height=20]file:///C:/Users/burgesss/AppData/Local/Temp/msohtmlclip1/01/clip_image004.png[/img] [12, 19]:[br][br]  x[sup]2[/sup] =[math]\frac{-1\pm\sqrt{1+4x}}{2}\ge0[/math] and   x[sup]2[/sup] = [math]\frac{-1\pm\sqrt{1+4x}}{2}\le0[/math] (to o ensure we have real values).[br]Solving the inequalities for x, we get:  [math]-4\le x\le-3[/math] and [math]3\le x\le4[/math].[br][br]Therefore, the pre-image of [12, 19] under [i]f[/i][sup]-1 [/sup]is the union of the intervals [-4, -3] and [3, 4]. denoted as [i]f[/i][sup]--1[/sup][12, 19] = [-4, -3] [math]\cup[/math] [3, 4]. [br][br]In summary, for the function f: Z [math]\rightarrow[/math] Z defined as f(x) = x[sup]2[/sup] + 3, we have: [i]f[/i][-3,5]) = [12, 28]  [br]Image of the interval [-3, 5].[i]f[/i][sup]--1[/sup][12, 19]) = [-4, -3] [math]\cup[/math] [3, 4] [br]Pre-image of the interval [12, 19] under the inverse function [i]f[/i][sup]-1[/sup].[/size]