C3 inverse functions question

Q5 of exercise 2E is a bit weird:
the function m(x) is defined by m(x)=x²+4x+9 {xEr,x>a} for 'some constant a'. if m^-1(x) exists, state the value of a and hence determine the eqn of m^-1(x).
well i can find the equation m-1(x) without a constant a; and how am i supposed to simply 'state the value of a' without any further info? if i solve the quadratic eqn x²+4x+9 for x, i'll be solving for x, not a. HOW am i to find a value for a? and how will that help in determining the eqn of m-1(x) anyway?
thanks.

Reply 1

the bear

20

it is to do with restricting the domain of m(x) so the remaining graph is one-to-one

Reply 2

james22

16

For a function to be invertable it has to be a bijection. A quadratic is never a bijection over the whole of R, but it is over a certain portion of it. If you draw the graph, look for the bit where it is bjective.

Reply 3

brianeverit

9

Original post by JacobAlevels

Q5 of exercise 2E is a bit weird:
the function m(x) is defined by m(x)=x²+4x+9 {xEr,x>a} for 'some constant a'. if m^-1(x) exists, state the value of a and hence determine the eqn of m^-1(x).
well i can find the equation m-1(x) without a constant a; and how am i supposed to simply 'state the value of a' without any further info? if i solve the quadratic eqn x²+4x+9 for x, i'll be solving for x, not a. HOW am i to find a value for a? and how will that help in determining the eqn of m-1(x) anyway?
thanks.

m(x)=(x+2)^2+5

so it will be a 1 - 1 function for x>-2, hence a=-2

C3 inverse functions question

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