Exercise 1.1.2[br][br]1. (a) Let [math] f: A\to B [/math] such that [math] f(x) = y [/math] and [math] g:B \to C [/math] such that g(y) = z. Name the function defined from A to C.[br]Solution:[br]The function defined from A to C is [math]gof(x) [/math].[br][br]1. (b) Write the difference between [math] (fog)(x) [/math] and [math] (gof)(x) [/math] .[br]Solution:[br] [math] (fog)(x) [/math] is the composite function of [math] g [/math] and [math]f [/math] and read as [math]g [/math] followed by [math]f [/math].[br][br] [math] (gof)(x) [/math] is the composite function of [math] f [/math] and [math]g [/math] and read as [math] f [/math] followed by [math] g [/math].[br][br]1. (c) Define composite function of [math] f [/math] and [math] g [/math].[br][br]Solution:[br]If [math] f:A \to B [/math] and [math] g:B\to C [/math] be any two functions then the new function devinded from [math] A \to C [/math] is called composite function of [math] f [/math] anf [math] g [/math]. It is denoted by [math] gof(x) [/math].[br][br]2. (a) Given [math]f [/math] and [math] g [/math] are two real-valued functions defined as below. Find (i) domain of [math] f [/math] (ii) domain of [math] g [/math] (iii) domain of [math] (fog) [/math] (iv) range of [math] (gof) [/math] in each of the following if exist.[br][br](a) [math] f = \{ (3,4),(5,6), (9,10) \} [/math] and [math] g =\{ (4,16),(6,36),(10,100) \} [/math].[br][br]Solution:[br](i) Domain of [math] f = \{ 3,5,9 \} [/math][br][br](ii) Domain of [math] g = \{ 4,6,10 \} [/math][br][br](iii) Since range of [math] g \ \ \cap [/math] domain of [math] f = \phi [/math] so, [math] fog [/math], does not exists.[br][br](iv) Range of [math] (gof) [/math] = range of [math] g = \{ 16,36,100 \} [/math][br][br](b) [math] f = \{(1, 2), (2, 3), (3, 4) \} [/math] and [math] g = \{(2, 3), (3, 4), (4, 5)\} [/math][br][br]Solution:[br][br](i) Domain of [math] f = \{ 1,2,3 \} [/math][br][br](ii) Domain of [math] g = \{ 2,3,4 \} [/math][br][br](iii) As range of [math] g [/math] is not a subset of domain of [math]f [/math] , so [math] fog [/math] can not be defined. Therefore, [math] fog [/math] does not exists.[br][br](iv) Range of [math] (gof) [/math] = Range of [math] g = \{ 3,4,5 \} [/math][br][br]3. Given that [math] f(x) = 2x-5 [/math] and [math] g(x) = x^2 -2x+6 [/math] , calculate:[br][br](a) [math] (fog) (x) [/math] and [math] (gof)(x) [/math][br][br]Solution:[br][math] \begin{align} [br](fog)(x) & = f(g(x)) \\[br]\ & = f( x^2 - 2x +6 ) \\[br]\ & = 2(x^2 - 2x +6) - 5 \\[br]\ & = 2x^2 -4x + 12 - 5 \\[br]\ & = 2x^2 -4x + 7[br] \end{align} [/math][br]Also,[br][math] \begin{align} [br](gof)(x) & = g(f(x))\\[br]\ & = g(2x-5) \\[br]\ & = (2x-5)^2 - 2(2x-5) +6 \\[br]\ & = (2x)^2 - 2\times 2x \times 5 + 5^2 -4x+10+6 \\[br]\ & = 4x^2 - 20 x + 25 -4x + 10 +6 \\[br]\ & = 4x^2 -24x+41 [br] \end{align} [/math][br][br](b) [math] (fog)(5) anf (gof)(4) [/math][br][br]Solution:[br][math] \begin{align} [br] (fog)(5) & = f(g(5)) \\[br]\ & = f( 5^2 - 2\times 5 +6 ) \\[br]\ & = f(25 -10+6) \\[br]\ & = f(21) \\[br]\ & = 2 \times 21 - 5 \\[br]\ & = 42 - 5 \\[br]\ & = 37 [br] \end{align} [/math][br]Also,[br][math] \begin{align} [br](gof)(4) & = g(f(4)) \\[br]\ & = g( 2 \times 4 - 5 ) \\[br]\ & = g(8 - 5 ) \\[br]\ & = g(3) \\[br]\ & = 3^2 - 2\times 3 +6 \\[br]\ & = 9 - 6 +6 \\[br]\ & = 9 \\[br] \end{align} [/math][br][br](c) [math] (gog)(2) [/math] and [math] (fof)(9) [/math][br][br]Solution:[br][math] \begin{align} [br] (gog)(2) & = g(g(2)) \\[br]\ & = g( 2^2 - 2\times 2 + 6 ) \\[br]\ & = g(4 - 4 +6 ) \\[br]\ & = g(6) \\[br]\ & = 6^2 -2\times 6 + 6 \\[br]\ & = 36 - 12 + 6 \\[br]\ & = 30\\[br] \end{align} [/math][br]Also,[br][math] \begin{align} [br](fof)(9) & = f(f(9)) \\[br]\ & = f( 2\times 9 - 5 ) \\[br]\ & = f( 18 - 5 ) \\[br]\ & = f(13 ) \\[br]\ & = 2\times 13 - 5 \\[br]\ & = 26 - 5 \\[br]\ & = 21 \\[br] \end{align} [/math][br][br](d) [math] (fog)(-4) [/math] and [math] (gof)(-4) [/math] [br][br]Solution:[br][math] \begin{align} [br](fog)(-4) & = f(g(-4)) \\[br]\ & = f( (-4)^2 - 2(-4) +6 ) \\[br]\ & = f( 16 +8 +6 ) \\[br]\ & = f(30) \\[br]\ & = 2\times 30 - 5 \\[br]\ & = 60 - 5 \\[br]\ & = 55 \\[br] \end{align} [/math][br]Also,[br][math] \begin{align} [br](gof)(-4) & = g(f(-4)) \\[br]\ & = g( 2\times (-4) -5 ) \\[br]\ & = g(-8 -5 ) \\[br]\ & = g(-13) \\[br]\ & = (-13)^2 - 2(-13) +6 \\[br]\ & = 169 + 26 + 6 \\[br]\ & = 201 \\[br] \end{align} [/math][br][br]4. (a) If [math] f(x) = x, g(x) = x + 1 [/math] and [math] h(x) = 2x – 1 [/math] then find [math] (f(goh)(x)) [/math] and [math] (go(hof)(x)) [/math].[br][br]Solution:[br][math] \begin{align} [br]\text{Given}, & \\[br]\ & f(x) = x \\[br]\ & g(x) = x+1 \\[br]\ & h(x) = 2x - 1 \\[br]\text{Now}& \\[br](goh)(x) & = g(h(x)) \\[br]\ & = g(2x-1) \\[br]\ & = 2x-1+1 \\[br]\ & = 2x \\[br]\therefore (goh)(x) & = 2x \\[br]\text{Also} & \\[br](f(goh)(x)) & = f( (gof)(x) ) \\[br]\ & = f(2x) \\[br]\ & = 2x \\[br]\therefore (f(goh)(x)) & = 2x \\[br] \end{align} [/math][br]Also,[br][math] \begin{align} [br](hof)(x) & = h(f(x)) \\[br]\ & = h(x) \\[br]\ & = 2x-1 \\[br]\therefore (hof)(x) & = 2x-1 \\[br]\text{Now} & \\[br](go(hof)(x)) & = g(hof(x)) \\[br]\ & = g( 2x-1) \\[br]\ & = 2x-1+1 \\[br]\ & = 2x \\[br](go(hof)(x)) & = 2x \\[br] \end{align} [/math][br][br]4. (b) Given that [math] f(x) = 2x -3, g(x) = x^3 + 2 [/math] and [math] h(x) = x^2 -2x+3 [/math], find [math] (f_o(g_oh)(x)) [/math] and [math] ((h_of)_og(x)) [/math]. (Taking composition of two functions as a single function).[br][br]Solution:[br][math] \begin{align} [br]\text{Given}, & \\[br]f(x) & = 2x -3 \\[br]g(x) & = x^3 + 2 \\[br]h(x) & = x^2 -2x+3 \\[br]\text{Now}& \\[br](g_oh)(x) & = g(h(x)) \\[br]\ & = g(x^2 - 2x +3 ) \\[br]\ & = (x^2 - 2x + 3)^3 + 2 \\[br]\text {Again,} & \\[br](f_o(g_oh)(x)) & = f(g_oh(x)) \\[br]\ & = f ( (x^2 - 2x +3 )^3 + 2 ) \\[br]\ & = 2 ((x^2 -2x+3)^3 +2) -3 \\[br]\ & = 2(x^2 -2x+3)^3 + 4 - 3 \\[br]\ & = 2(x^2 -2x+3)^3 + 1 \\[br]\therefore(f_o(g_oh)(x)) & = 2(x^2 -2x+3)^3 +1 \\[br]\text{Also,} & \\[br] (h_of)(x) & = h (f(x)) \\[br]\ & = h( 2x-3) \\[br]\ & = (2x-3)^2 - 2(2x-3)+3 \\[br]\ & = 4x^2 - 12x+9 - 4x+6 + 3 \\[br]\ & = 4x^2 - 16x + 18 \\[br]\therefore (h_o f)(x) & = 4x^2 - 16x + 18 \\[br]\text{Now,} & \\[br]((h_of)_og(x)) & = (h_o f)(g(x)) \\[br]\ & = (h_o f)( x^3 +2 ) \\[br]\ & = 4(x^3+2)^2-16(x^3 + 2) + 18 \\[br]\therefore ((h_of)_og(x)) & = 4(x^3+2)^2-16(x^3 + 2) + 18 \\[br] \end{align} [/math][br][br]5. If [math] f [/math] and [math] g [/math] are linear functions, what can you say about the domain of [math](f_o g) [/math] and [math] (g_o f) [/math] ? Explain.[br][br]Solution:[br][br]The domain of [math](f_o g) [/math] = Doman of [math] g = - \infty < x < \infty [/math].[br][br]The domain of [math] (g_o f) [/math] = Domain of [math] f = - \infty < x < \infty [/math].[br][br]6. Dolma determines the domain of [math] f_o g [/math] by examining only the formula for [math] (f_o g)(x). [/math] Is her approach valid? Why or why not?[br]Solution:[br][br]If Dolma determines the domain of [math] fog [/math] by examining only the formula of [math] (fog)(x) [/math] then her approach may not be valid. We know the domain of [math] fog(x) [/math] is same as the domain of [math] g(x). [/math] If formula of [math] g(x) [/math] contains radical sign or fraction having zero values in the denominator for some values of [math] x [/math] then the domain of [math] fog(x) [/math] cannot be examined just by the formula of [math] (fog)(x). [/math][br]For example,[br][br](i)Let’s define the functions [math] f [/math] an [math] g [/math] by [math] f (x) = \frac{1 }{x} [/math] and [math] g (x) = \frac{1}{x^2} [/math] then [math] (fog)(x) = x^2 [/math] which is defined at [math] x = 0 [/math] but [math] g (x) [/math] is not defined at [math] x = 0. [/math] That means 0 is not the element in the domain of [math] g [/math] and hence 0 is not the element in the domain of [math] fog [/math].[br](ii)Let’s define the functions [math] f [/math] an [math] g [/math] by [math] f (x) = \frac{1}{x-2} [/math] and [math] g (x) = \frac{1}{ x-3 } [/math] then [math] (fog)(x) = \frac{ x-3}{4-x } [/math] which is defined at [math] x = 3 [/math] but [math] g (x) [/math] is not defined at [math] x = 3. [/math] That means 3 is not the element in the domain of g and hence 3 is not the element in the domain of [math] (fog)(x). [/math][br]From the above counter examples, we can claim that her approach is not valid.[br][br]Her approach is valid if and only if both [math] f(x) [/math] and [math] g(x) [/math] and are linear functions.[br][br]Alternative:[br]Let us start from an example.[br][math] f(x) = \frac{1}{x} \ \ x\neq 0 [/math] [br][br] [math] g(x) = \frac{1}{x-1} \ \ x\neq 1 [/math][br][br]Then, [br][math] \begin{align} fog(x) & = f(g(x)) \\ & = f \left( \frac{1}{x-1} \right) \\ & = x - 1 \\ \therefore fog(x) & = x - 1 \end{align} [/math] [br][br]Now if we examine domain of [math] fog [/math] just by considering its formula for [math] fog(x) [/math], certainly it will be whole number [math] R [/math], also range will again be whole number [math] R [/math]. But as range for [math] f [/math] and [math] g [/math] excludes certain points in [math] R [/math], so will [math] fog [/math]. Hence it will not be valid to examine domain of a composite function just by taking its direct formula into account. In example above, domain for [math] fog [/math] must be certainly [math] R - \{ 0, 1 \} [/math].[br][br]7. Write yourself any two real-valued function. Find their composition.[br]Solution:[br]Let [math] f(x) = 2x - 1 \text{ and } g(x) = 3 - 5x [/math][br]Now,[br][math] \begin{align} fog(x) & = f(g(x)) \\ & = f( 3-5x) \\ & = 2(3-5x) - 1 \\ & = 6 - 10x - 1 \\ & = 5 - 10x \\ \therefore fog(x) & = 5 - 10x \end{align} [/math] [br]Also, [br][math] \begin{align} gof(x) & = g(f(x)) \\ & = g(2x-1) \\ & = 3-5(2x-1) \\ & = 3 - 10x+5 \\ & = 8 -10x \\ \therefore gof(x) & = 8 - 10x \end{align} [/math] [br][br]8. A stone is thrown into a pond, creating a circular ripple that spreads over the pond in [br]such a way that the radius is increasing at the rate of 3 ft/sec.[br](a) Find a function [math] r(t) [/math] for the radius in terms of [math] t. [/math][br]Solution:[br] [math] r(t) = 3t [/math][br](b) Find a function [math] A(r) [/math] for the area of the ripple in terms of the radius [math] r.[/math][br]Solution:[br] [math] A(r) = \pi r^2 [/math] [br](c) Find [math] (A_o r)(t). [/math] Explain the meaning of this function.[br]Solution:[br][math] \begin{align} [br]( A_o r ) (t) & = A(r(t)) \\[br]\ & = A ( 3t ) \\[br]\ & = \pi (3t)^2 \\[br]\ & = \pi (9 t^2 ) \\[br]\ & = 9 \pi t^2 \\[br]\therefore ( A_o r ) (t) & = 9 \pi t^2 \\[br] \end{align} [/math][br]This function represents the area of circular ripple after [math] t [/math] second.