Functions

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Dec 14th 2009, 02:47 AM
EmG

Functions

Anyone know this one?

$f(x)=9x^3-48x^2+64x$
Dec 14th 2009, 02:54 AM
Raoh

Quote:

Originally Posted by EmG

Anyone know this one?

$f(x)=9x^3-48x^2+64x$

hello
factorize with x.
$x(9x^{2}-48x+64)=x(8-3x)^{2}$
Dec 14th 2009, 02:55 AM
EmG

I did and when I did it I get stuck at $f(x)=x(9x^2-48x+64)$
Dec 14th 2009, 02:58 AM
Raoh

Quote:

Originally Posted by EmG

I did and when I did it I get stuck at $f(x)=x(9x^2-48x+64)$

$9x^2-48x+64=(3x)^2-48x+8^2$
is it clear now ?
(Happy)
Dec 14th 2009, 03:00 AM
Raoh

Remember !
$(a-b)^2=a^2-2ab+b^2$
(Happy)
Dec 14th 2009, 03:02 AM
EmG

Ummm.. Sadly I do not understand it very well or how to get the zeros from it algebraically.
Dec 14th 2009, 03:08 AM
Raoh

Quote:

Originally Posted by EmG

Ummm.. Sadly I do not understand it very well or how to get the zeros from it algebraically.

okay..
do you agree that $f(x)=x(8-3x)^{2}$
if yes,all you have to do is to put f(x)=0 and find the solutions.
Dec 14th 2009, 03:16 AM
EmG

When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?
Dec 14th 2009, 03:22 AM
Raoh

Quote:

Originally Posted by EmG

When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?

$f(x)=0\Leftrightarrow x(8-3x^2)=0$
therefore the solutions are $\left [ x=0,x=\frac{8}{3} \right ]$ .
hope that helps.
Dec 14th 2009, 04:10 AM
mr fantastic

Quote:

Originally Posted by EmG

When I pluged it in I got 0=9 so would that mean that there is no zeros in the equation?

Go back and review the basic theory that is undoubtedly given in your classnotes or textbook. You're expected to know that

$f(x)=0 \Rightarrow x(8-3x^2)=0 \Rightarrow x = 0 \, \text{or} \, 8 - 3x^2 = 0 \Rightarrow x = ....$

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