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Inverse Function : Problems and Solutions

1.       Given that f(x)=ex+3 where xR, find f1(x) and state the domain of f1. Hence solve the equation f1(x)=ln(1x).

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          f(x)=ex+3, xRf1(y)=xf(x)=yex+3=yx+3=logeyx+3=lnyx=lny3f1(y)=lny3f1(x)=lnx3Domain of f1={x|x>0, xR}f1(x)=ln(1x)lnx3=ln(1x)lnxln(1x)=3lnx2=32lnx=3lnx=32x=e32


2.        A function f is defined by f(3x2)=5+6x. Find the value of f1(29).

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          f(3x2)=5+6xf1(5+6x)=3x2Let 5+6x=29,then6x=24x=4f1(29)=3(4)2=10


3.        A function f is defined by f(x)=x32x5.

(i) State the value of x for which f is not defined.

(ii) Find the value of x for which f(x)=0.

(iii) Find the inverse functionf1 and state the domain of f1.

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                 f(x)=x32x5

         (i)    f is not defined when

                2x5=0x=52

         (ii)   f(x)=0

                x32x5=0

                Since 2x50,

                x3=0x=3

         (iii)  Let f1(x)=y, then

                 f(y)=x

                y32y5=x

                y3=2xy5x       y2xy=35x       y(12x)=35x

                y=35x12x, x12

                Domain of f1={x|xR,x12}


4.       A function f is defined by f:xax+1, x0 where a is a constant. Given that 6(ff)(1)+f1(2)=0, find the possible values of a.

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         f(x)=ax+1, x0

         (ff)(1)=f(f(1))

                         =f(a1+1)

                         =f(1a)

                         =a1a+1

                         =11a

         f1(2)=kf(k)=2

         ak+1=2

            ak=1k=a

         f1(2)=a

         6(ff)(1)+f1(2)=0

         61a+a=0

            6+aa2=0

         (3a)(2+a)=0

         a=3 or a=2


5.       A function g is defined by g:xx+1x2, x2,x5 and h is defined by is defined by h:xax+3x, x0. Given that (hg1)(4)=6, calculate the value of a.

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                g(x)=x+1x2, x2,x5

                h(x)=ax+3x, x0

                (hg1)(4)=6      g1(4)=pg(p)=4

              p+1p2=4

                p+1=4p8      3p=9p=3    g1(4)=3      (hg1)(4)=6      h(g1(4))=6      h(3)=6

                a(3)+33=6

              a+1=6a=5


6.       Let f:RR and g:RR be defined by f(x)=3x1 and g(x)=x+7. Find (f1g)(x) and what is the value of aR for which (f1g)(a)=3.

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         f:RR, f(x)=3x1g:RR, g(x)=x+7(f1g)(x)=f1(g(x))Let f1(g(x))=y then g(x)=f(y).x+7=3y1

            y=x+83

         (f1g)(x)=x+83

            (f1g)(a)=3

            a+83=3a=1


7.       For the function f(x)=2x3x+1, x13 find f1 and verify that (ff1) and (f1f) both equal I.

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              f(x)=2x3x+1, x13

              f1(x)=yf(y)=x     2y3y+1=x

              2y=3xy+x    y(23x)=x

              y=x23x

            f1(x)=x23x, x23

            (ff1)(x)=f(f1(x))

                                 =f(x23x)

                                 =2x23x3x23x+1

                                 =2x23x×23x3x+23x

                                 =x                       =I(x)  (f1f)(x)=f1((x))

                                 =f1(2x3x+1)

                                 =2x3x+126x3x+1

                                 =2x3x+1×3x+16x+26x

                                 =x                       =I(x)  (ff1)(x)=(f1f)(x)=I(x)


8.       Functions f and g are defined, for xR, by f:x5x2,g:x12x1, x12. Find the value of x for which

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         (i) f(x)=f1(x).

         (ii) (fg)(x)+3g(x)=0.

                   f(x)=5x2,

                   g(x)=12x1, x12

         (i)       f(x)=f1(x)        f(f(x))=x        f(5x2)=x        5(5x2)2=x        24x=12

                 x=12

         (ii)      (fg)(x)+3g(x)=0          f(g(x))+3g(x)=0

                   f(12x1)+32x1=0

                   52x12+32x1=0

                 82x1=2

                 2x1=4

                x=52


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