Question

What is the inverse of g(x) = x^2-5x-14 Help please :)

Answer 1

u mean its inverse function?

Answer 2

it is not a one to one function, it does not have an inverse
however you can still solve
\[x=y^2+5y-14\] for \(y\) if you like

Answer 3

I have to restrict the domain and range so I can find the inverse....doing my head in LOL

Answer 4

ok.

Answer 5

so u have
\[\large y=x^2-5x-14 \]
\[\large y+14+25/4=x^2-5x+\frac{25}{4} \]
\[\large y+81/4=(x-5/2)^2 \]

Answer 6

this shows that the function is symetric in respect with x=5/2. therefore u can restrict the domain to either \([5/2;\infty)\) or \((-\infty;5/2]\).

Answer 7

sorry *with respect to

Answer 8

where does 25/4 come from?

Answer 9

from completing the square

Answer 10

u still have to solve for \(x\) the equation
\[\large y=f(x)=x^2-5x-14 \]

Answer 11

so the inverse is y+81/4=(x-5/2)^2?

Answer 12

no. that was just some calculations to find out the restriction of the domain. u have to solve the equation
\[\large y=f(x)=x^2-5x-14 \]
for \(x\).

Answer 13

x is 7 or -2

Answer 14

no. u have basicly a quadratic equation for x:
\[\large x^2-5x-(14+y)=0 \]
using the quadratic formula u get
\[\large x=\frac{5\pm\sqrt{(-5)^2-4\cdot1\cdot(-(14+y))}}{2\cdot1} \]

Answer 15

\[\large x=\frac{5\pm\sqrt{81+4y}}{2} \]

Answer 16

so the inverse function would be either
\[\large f^{-1}(y)=\frac{5+\sqrt{81+4y}}{2} \]
or
\[\large f^{-1}(y)=\frac{5-\sqrt{81+4y}}{2} \]
deppending on which interval u chose for the domain.

Answer 17

x\[x \ge-\frac{ 81 }{ 4 } y \ge \frac{ 5 }{ 2 }\]

Answer 18

domain corresponds to \(x\) and the range to \(y\).

Answer 19

so which inverse function would I choose?

Answer 20

if u choose \([5/2;\infty)\) for the domain then
\[\large f^{-1}(x)=\frac{5+\sqrt{81+4x}}{2} \]

Answer 21

Thank you so very very much for all of your help, I really appreciate it :)

Answer 22

u r welcome

Answer 23

so I just sub in 5/2 in the equation and see what I get as to whether it satisfies the domain ?

Answer 24

NO. the domain is the set of points (real numbers) for which the equation (function) makes sense. since your function has no restrictions (logarithms, divisions, or roots of even power) its domain is actually the whole of \(\mathbb{R}\).

BUT. since the function is not one-to-one on this domain, u were told to "reduce" its domain to a set upon which the function is one-to-one.