functions differenciate

**brokedude** · August 23rd 2007, 05:34 PM

im terrible @ math. please help me with this!

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...

**CaptainBlack** · August 23rd 2007, 07:42 PM

Originally Posted by brokedude

im terrible @ math. please help me with this!

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:

$[g(f)](x)=g(f(x))=-2(f(x))^2+5f(x)=2(2x)^2+5(2x)=8x^2+10x<br />$

Second part:

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

RonL

**CaptainBlack** · August 23rd 2007, 07:47 PM

Originally Posted by brokedude

im terrible @ math. please help me with this!

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 $f$ is:

$<br /> DQ(f,h)=\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)-2x}{h}=\frac{2h}{h}=2<br />$

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

RonL

**topsquark** · August 24th 2007, 04:26 AM

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

Originally Posted by CaptainBlack

First part:

Second part:

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

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:
$g(x) = -2x^2 + 5x$

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

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

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

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

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

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

**brokedude** · August 24th 2007, 11:38 AM

thanks for your help !!

**CaptainBlack** · August 24th 2007, 01:28 PM

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:
$g(x) = -2x^2 + 5x$

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

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

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

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

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

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 $g$ from $\bold{R}$ to $\mathcal{P}(\bold{R})$ , when:

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

but I doubt that this is what is wanted.

RonL

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