# functions differenciate

• Aug 23rd 2007, 05:34 PM
brokedude
functions differenciate

given 2 functions: f(x) = 2x, g(x) = -2x²+5x

im supposed to determine the following

g(f)
inverse of g
g (g inverse)
the difference quotient for f
the average rate of change for f

yeah... a lot of Q's. someone pls explain the process to me, cuz i fail to understand it...
• Aug 23rd 2007, 07:42 PM
CaptainBlack
Quote:

Originally Posted by brokedude

given 2 functions: f(x) = 2x, g(x) = -2x²+5x

im supposed to determine the following

g(f)
inverse of g
g (g inverse)

First part:

$\displaystyle [g(f)](x)=g(f(x))=-2(f(x))^2+5f(x)=2(2x)^2+5(2x)=8x^2+10x$

Second part:

There is no function $\displaystyle g^{-1}$, as for most $\displaystyle y$ there are two $\displaystyle x$s such that $\displaystyle g(x)=y$.

RonL
• Aug 23rd 2007, 07:47 PM
CaptainBlack
Quote:

Originally Posted by brokedude

given 2 functions: f(x) = 2x, g(x) = -2x²+5x

im supposed to determine the following:

the difference quotient for f
the average rate of change for f

The difference quotient for $\displaystyle f$ is:

$\displaystyle DQ(f,h)=\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)-2x}{h}=\frac{2h}{h}=2$

Thus we see that the average rate of change of $\displaystyle f$ over
any interval $\displaystyle (x, x+h)$ is $\displaystyle 2$

RonL
• Aug 24th 2007, 04:26 AM
topsquark
Quote:

Originally Posted by brokedude
g(x) = -2x²+5x

im supposed to determine the following

g(f)
inverse of g
g (g inverse)
the difference quotient for f
the average rate of change for f

Quote:

Originally Posted by CaptainBlack
First part:

Second part:

There is no function $\displaystyle g^{-1}$, as for most $\displaystyle y$ there are two $\displaystyle x$s such that $\displaystyle g(x)=y$.

RonL

CaptainBlack is absolutely correct about the inverse of g. It has no inverse. However we may informally get an inverse (that is to say, the process is correct even if the application is not) by:
$\displaystyle g(x) = -2x^2 + 5x$

Let $\displaystyle y = -2x^2 + 5x$

Now switch the roles of x and y:
$\displaystyle x = -2y^2 + 5y$

Now solve for y:
$\displaystyle 2y^2 - 5y + x = 0$

$\displaystyle y = \frac{5 \pm \sqrt{25 - 8x}}{4}$ <-- via the quadratic formula

Thus
$\displaystyle g^{-1}(x) = \frac{5 \pm \sqrt{25 - 8x}}{4}$

The problem, as CaptainBlack mentioned, is that for the graph y = g(x) there are two y values for every x (except at the vertex point), so we need to be very careful about defining a domain on which an inverse exists and just what that inverse is. (It will either be the "+" or the "-" of the inverse formula given above.)

-Dan
• Aug 24th 2007, 11:38 AM
brokedude
• Aug 24th 2007, 01:28 PM
CaptainBlack
Quote:

Originally Posted by topsquark
CaptainBlack is absolutely correct about the inverse of g. It has no inverse. However we may informally get an inverse (that is to say, the process is correct even if the application is not) by:
$\displaystyle g(x) = -2x^2 + 5x$

Let $\displaystyle y = -2x^2 + 5x$

Now switch the roles of x and y:
$\displaystyle x = -2y^2 + 5y$

Now solve for y:
$\displaystyle 2y^2 - 5y + x = 0$

$\displaystyle y = \frac{5 \pm \sqrt{25 - 8x}}{4}$ <-- via the quadratic formula

Thus
$\displaystyle g^{-1}(x) = \frac{5 \pm \sqrt{25 - 8x}}{4}$

The problem, as CaptainBlack mentioned, is that for the graph y = g(x) there are two y values for every x (except at the vertex point), so we need to be very careful about defining a domain on which an inverse exists and just what that inverse is. (It will either be the "+" or the "-" of the inverse formula given above.)

-Dan

There is a way around the problem which is to extend $\displaystyle g$ from $\displaystyle \bold{R}$ to $\displaystyle \mathcal{P}(\bold{R})$, when:

$\displaystyle g(g^{-1})=I$ the identity function $\displaystyle I(S)=S$ for all subsets of $\displaystyle \bold{R}$

but I doubt that this is what is wanted.

RonL