This document provides explanations and examples on finding the domain and range of functions. It covers various types of functions and their properties. Practice problems are included to enhance understanding.
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Solution: 1(f):Given that Note that domain of a function is the set of input values for which the function is real and defined. Sowill be defined and real ifand Now solve, this gives. Since, so discard. So, the domain ofis the set of all values exceptand In interval notation, the domain ofis. Solution: 1(g)Given that Note that domain of a function is the set of input values for which the function is real and defined. Sowill be defined and real if. Sincethis implies thatfor real. Hence domain ofis Solution 2(d):Given So, Now, Hence Solution 2(f):Given. Now, Hence, Solution 3(b):Given. Since,, so g(x) is within f(x). Hencesuch that h(x) = (f o g) (x). Solution 3(d):Given.
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Since,, so g(x) is within f(x). Hencesuch that h(x) = (f o g) (x). Solution 4(b):Given. So, Hence, Solution 4(c):Given Since,. So, Hence, Solution 5(a):Given. Letthen, Solve for: This implies that. Replace y with x we get, Sinceisdefinedandrealforeveryxexceptx=5,hencedomainis . Sinceis defined for everyxexceptx = 1, hence range of is. Now,
Hence, Solution 8(a):Given. The formula for difference quotient is. Now, So, the difference quotient is Hence, Solution 8(b):Given. The formula for difference quotient is. Now, . So, the difference quotient is Hence, Solution 11(a):Given. Since f(x) passes through the points So,and Divide equation (i) by equation (ii) we get, Substitute the value of b in equation (ii) we get,
Hence, And the graph of f(x) is shown below: From the graph, it is observe that it represent exponential decay. Solution 12(c):Given. To find y - intercept, substitute t = 0 we get, y = g(0) = 4 + 5 = 9 And the graph of the exponential function is shown below: Solution 14(a):To simplify the expression. Now, Hence, Solution 14(b):To simplify the expression. Now,
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Hence, Solution 21:Given, (a): (b):From the above graph values of y varies fromto 1. Hence range of the given function is. (c):From the graph, when x goes to, f goes toand when x goes to, f goes to 0. (d):From the graph, the function is increasing over the intervaland decreasing over the interval. Solution 22(i):Given (a):For y β intercept, substitute x = 0, we get. Hence y β intercept is And for x β intercept, substitute y = 0, we get,. Hence x β intercepts are (b):Vertical asymptote: Note that for a rational function, the vertical asymptotes are the zeros of the denominator. Since y is undefined at x = 3 and x = -3 so, x = 3 and x = -3 are the vertical asymptote. Horizontal asymptote: Since degree of numerator and denominator are same. So the horizontal asymptote is Hence, the horizontal asymptote is y = 1
Solution 22(ii):Given (a):For y β intercept, substitute x = 0, we get. Hence y β intercept is And for x β intercept, substitute y = 0, we get,. Hence x β intercepts are (b):Vertical asymptote is Since, numerator degree > 1 + denominator degree. Hence there is not horizontal asymptote. Solution 22(iii):Given (a):For y β intercept, substitute x = 0, we get. Hence y β intercept is And for x β intercept, substitute y = 0, we get,. Hence x β intercepts are (b):Vertical asymptote: Note that for a rational function, the vertical asymptotes are the zeros of the denominator. Since y is undefined at x = -2 and x = 3 so, x = -2 and x = 3 are the vertical asymptote. Horizontal asymptote: Since degree of numerator and denominator are same. So the horizontal asymptote is Hence, the horizontal asymptote is y = -3 Solution 22(iv):Given (a):For y β intercept, substitute x = 0, we get. Hence y β intercept is And for x β intercept, substitute y = 0, we get,, an imaginary number Hence, there is no x βintercept. (b):Vertical asymptote: Note that for a rational function, the vertical asymptotes are the zeros of the denominator. Since y is undefined at x = 1 and x = 4 so, x = 1 and x = 4 are the vertical asymptote. Horizontal asymptote: Since degree of numerator and denominator are same. So the horizontal asymptote is Hence, the horizontal asymptote is y = 1