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Graphs - Inverse Functions

  a - answer    s - solution    v - video

Question 1


Find $f(x)$

a) $f(x)=2x$ when $x=3$  a s v
b) $f(x)=\frac x5$ when $x=20$  a s v
c) $f(x)=x+16$ when $x=12$  a s v
d) $f(x)=x-3$ when $x=10$  a s v
e) $f(x)=3x+11$ when $x=4$  a s v
f) $f(x)=\frac x6-2$ when $x=18$  a s v
g) $f(x)=x^2+5$ when $x=2$  a s v
h) $f(x)=3-\sqrt x$ when $x=36$  a s v


Question 2


Find the inverse function for the following functions

a) $f(x)=x+2$  a s v
b) $g(x)=x-5$  a s v
c) $h(x)=3x$  a s v
d) $p(x)=x\div6$  a s v
e) $q(x)=\frac x9$  a s v
f) $f(y)=y+t$  a s v
g) $g(a)=\frac ab$  a s v
h) $h(z)=\sqrt z$  a s v
i) $p(x)=x^5$  a s v
j) $m(r)=-r$  a s v
k) $f(x)=\frac1x$  a s v


Question 3


Show that the following are the inverse when $x=4$

a) $f(x)=x^3$ and $f^{-1}(x)=\sqrt[3] x$  a s v
b) $f(x)=2+3x$ and $f^{-1}(x)=\frac{x-2}{3}$  a s v
c) $f(x)=\frac{2+x}{4}$ and $f^{-1}(x)=4x-2$  a s v
d) $g(x)=\sqrt{x+12}-3$ and $g^{-1}(x)={(x+3)}^2-12$  a s v
e) $h(x)=\sqrt{x+12}-3$ and $h^{-1}(x)={(x+3)}^2-12$  a s v
f) $q(x)=\frac1x$ and $q^{-1}(x)=\frac1x$  a s v
g) $p(x)=\frac14x-7$ and $p^{-1}(x)=4(x+7)$  a s v
h) $f(x)=\frac{(2x+3)^2}{11}$ and $f^{-1}(x)=\frac{\sqrt{11x}-3}{2}$  a s v


Question 4


Find the inverse function for the following functions

a) $f(x)=2x+3$  a s v
b) $f(x)=-5x+6$  a s v
c) $f(x)=\frac12x-3$  a s v
d) $g(x)=\frac15x+\frac23$  a s v
e) $f(x)=\frac{x+1}{4}$  a s v
f) $n(x)=\frac{x-2}{-5}$  a s v
g) $f(a)=\sqrt a+15$  a s v
h) $g(y)={(y+3)}^3$  a s v


Question 5


By showing that $f(f^{-1}(x))=x$ prove that the following inverse functions are given:

a) $f(x)=2x$ and $f^{-1}(x)=\frac x2$  a s v
b) $f(x)=\frac14x$ and $f^{-1}(x)=4x$  a s v
c) $f(x)=4x-1$ and $f^{-1}(x)=\frac{x+1}{4}$  a s v
d) $f(x)=\sqrt{5x}+4$ and $f^{-1}(x)=\frac{{(x-4)}^2}{5}$  a s v
e) $f(x)=\frac{x+4}{2x-5}$ and $f^{-1}(x)=\frac{5x+4}{2x-1}$  a s v