1. Given that f(x)=ex+3 where x∈R, find f−1(x) and state the domain of f−1. Hence solve the equation f−1(x)=ln(1x).
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f(x)=ex+3, x∈Rf−1(y)=x⇔f(x)=y∴ex+3=y∴x+3=logey∴x+3=lny∴x=lny−3∴f−1(y)=lny−3∴f−1(x)=lnx−3Domain of f−1={x|x>0, x∈R}f−1(x)=ln(1x)lnx−3=ln(1x)lnx−ln(1x)=3lnx2=32lnx=3lnx=32x=e32
2. A function f is defined by f(3x−2)=5+6x. Find the value of f−1(29).
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f(3x−2)=5+6x∴f−1(5+6x)=3x−2Let 5+6x=29,then6x=24⇒x=4∴f−1(29)=3(4)−2=10
3. A function f is defined by f(x)=x−32x−5.
(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 functionf−1 and state the domain of f−1.
(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 functionf−1 and state the domain of f−1.
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f(x)=x−32x−5
(i) f is not defined when
2x−5=0⇒x=52
(ii) f(x)=0
x−32x−5=0
Since 2x−5≠0,
x−3=0⇒x=3
(iii) Let f−1(x)=y, then
f(y)=x
y−32y−5=x
y−3=2xy−5x y−2xy=3−5x y(1−2x)=3−5x
y=3−5x1−2x, x≠12
Domain of f−1={x|x∈R,x≠12}
(i) f is not defined when
2x−5=0⇒x=52
(ii) f(x)=0
x−32x−5=0
Since 2x−5≠0,
x−3=0⇒x=3
(iii) Let f−1(x)=y, then
f(y)=x
y−32y−5=x
y−3=2xy−5x y−2xy=3−5x y(1−2x)=3−5x
y=3−5x1−2x, x≠12
Domain of f−1={x|x∈R,x≠12}
4. A function f is defined by f:x↦ax+1, x≠0 where a is a constant. Given that 6(f⋅f)(−1)+f−1(2)=0, find the possible values of a.
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f(x)=ax+1, x≠0
(f⋅f)(−1)=f(f(−1))
=f(a−1+1)
=f(1−a)
=a1−a+1
=11−a
f−1(2)=k⇔f(k)=2
∴ak+1=2
ak=1⇒k=a
∴f−1(2)=a
6(f⋅f)(−1)+f−1(2)=0
∴61−a+a=0
6+a−a2=0
∴(3−a)(2+a)=0
∴a=3 or a=−2
(f⋅f)(−1)=f(f(−1))
=f(a−1+1)
=f(1−a)
=a1−a+1
=11−a
f−1(2)=k⇔f(k)=2
∴ak+1=2
ak=1⇒k=a
∴f−1(2)=a
6(f⋅f)(−1)+f−1(2)=0
∴61−a+a=0
6+a−a2=0
∴(3−a)(2+a)=0
∴a=3 or a=−2
5. A function g is defined by g:x↦x+1x−2, x≠2,x≠5 and h is defined by is defined by h:x↦ax+3x, x≠0. Given that (h⋅g–1)(4)=6, calculate the value of a.
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g(x)=x+1x−2, x≠2,x≠5
h(x)=ax+3x, x≠0
(h⋅g−1)(4)=6 g−1(4)=p⇔g(p)=4
∴ p+1p−2=4
p+1=4p−8 3p=9⇒p=3∴ g−1(4)=3 (h⋅g−1)(4)=6 h(g−1(4))=6 h(3)=6
a(3)+33=6
∴ a+1=6⇒a=5
h(x)=ax+3x, x≠0
(h⋅g−1)(4)=6 g−1(4)=p⇔g(p)=4
∴ p+1p−2=4
p+1=4p−8 3p=9⇒p=3∴ g−1(4)=3 (h⋅g−1)(4)=6 h(g−1(4))=6 h(3)=6
a(3)+33=6
∴ a+1=6⇒a=5
6. Let f:R→R and g:R→R be defined by f(x)=3x−1 and g(x)=x+7. Find (f−1⋅g)(x) and what is the value of a∈R for which (f−1⋅g)(a)=3.
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f:R→R, f(x)=3x−1g:R→R, g(x)=x+7(f−1⋅g)(x)=f−1(g(x))Let f−1(g(x))=y then g(x)=f(y).∴x+7=3y−1
y=x+83
∴(f−1⋅g)(x)=x+83
(f−1⋅g)(a)=3
a+83=3⇒a=1
y=x+83
∴(f−1⋅g)(x)=x+83
(f−1⋅g)(a)=3
a+83=3⇒a=1
7. For the function f(x)=2x3x+1, x≠−13 find f−1 and verify that (f⋅f−1) and (f−1⋅f) both equal I.
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f(x)=2x3x+1, x≠−13
f−1(x)=y⇔f(y)=x 2y3y+1=x
2y=3xy+x y(2−3x)=x
y=x2−3x
∴ f−1(x)=x2−3x, x≠23
∴ (f⋅f−1)(x)=f(f−1(x))
=f(x2−3x)
=2x2−3x3x2−3x+1
=2x2−3x×2−3x3x+2−3x
=x =I(x)∴ (f−1⋅f)(x)=f−1((x))
=f−1(2x3x+1)
=2x3x+12−6x3x+1
=2x3x+1×3x+16x+2−6x
=x =I(x)∴ (f⋅f−1)(x)=(f−1⋅f)(x)=I(x)
f−1(x)=y⇔f(y)=x 2y3y+1=x
2y=3xy+x y(2−3x)=x
y=x2−3x
∴ f−1(x)=x2−3x, x≠23
∴ (f⋅f−1)(x)=f(f−1(x))
=f(x2−3x)
=2x2−3x3x2−3x+1
=2x2−3x×2−3x3x+2−3x
=x =I(x)∴ (f−1⋅f)(x)=f−1((x))
=f−1(2x3x+1)
=2x3x+12−6x3x+1
=2x3x+1×3x+16x+2−6x
=x =I(x)∴ (f⋅f−1)(x)=(f−1⋅f)(x)=I(x)
8. Functions f and g are defined, for x∈R, by f:x↦5x−2,g:x↦12x−1, x≠12. Find the value of x for which
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(i) f(x)=f−1(x).
(ii) (f⋅g)(x)+3g(x)=0.
f(x)=5x−2,
g(x)=12x−1, x≠12
(i) f(x)=f−1(x) ∴ f(f(x))=x ∴ f(5x−2)=x ∴ 5(5x−2)−2=x ∴ 24x=12
∴ x=12
(ii) (f⋅g)(x)+3g(x)=0 f(g(x))+3g(x)=0
f(12x−1)+32x−1=0
52x−1−2+32x−1=0
∴ 82x−1=2
∴ 2x−1=4
∴ x=52
(ii) (f⋅g)(x)+3g(x)=0.
f(x)=5x−2,
g(x)=12x−1, x≠12
(i) f(x)=f−1(x) ∴ f(f(x))=x ∴ f(5x−2)=x ∴ 5(5x−2)−2=x ∴ 24x=12
∴ x=12
(ii) (f⋅g)(x)+3g(x)=0 f(g(x))+3g(x)=0
f(12x−1)+32x−1=0
52x−1−2+32x−1=0
∴ 82x−1=2
∴ 2x−1=4
∴ x=52
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