I'm struggling with graphs. Any help appreciated. Thanks

The figure shows the graphs off(x)=x3 andg(x)=ax3. What can

you conclude about the value ofa?

a.a< –1

b . –1 <a< 0

c . 0 <a< 1

d . 1 <a

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- Jun 12th 2007, 05:38 PMwvmcanelly@cableone.netGraph HelpI'm struggling with graphs. Any help appreciated. Thanks

The figure shows the graphs of*f(x)*=*x*3 and*g(x)*=*ax*3. What can

you conclude about the value of*a*?

a.*a*< –1

b . –1 <*a*< 0

c . 0 <*a*< 1

d . 1 <*a*

- Jun 12th 2007, 07:08 PMThePerfectHacker
We see the following:

1)$\displaystyle f(x) = g(x) \mbox{ for }x=0$

2)$\displaystyle f(x) > g(x) \mbox{ for }x>0$

3)$\displaystyle f(x) < g(x) \mbox{ for }x<0$

Let us examine #2:

We know that $\displaystyle f(x) = x^3 \mbox{ and }g(x)=ax^3$

Thus,

$\displaystyle x^3 > ax^3 \mbox{ for }x>0$

$\displaystyle x^3 - ax^3>0$

$\displaystyle x^3 (1 - a) > 0$

Since $\displaystyle x>0 \Rightarrow x^3 >0$

Thus, $\displaystyle (1-a)>0 \Rightarrow a<1$.

Let us examine #3:

We know that $\displaystyle x^3 < ax^3 \mbox{ for } x<0$

$\displaystyle x^3 - ax^3 < 0$

$\displaystyle x^3 ( 1 - a) < 0$

This leads to, $\displaystyle a<1$ but we already know that, so we did not gain anything.

But if we look at $\displaystyle g(x)$ we see that $\displaystyle g(x)>0$ for $\displaystyle x>0$.

Thus,

$\displaystyle ax^3 \mbox{ for }x>0$

Since $\displaystyle x>0$ it tells us that $\displaystyle a>0$.

Thus,

$\displaystyle 0<a<1$. - Jun 12th 2007, 07:09 PMwvmcanelly@cableone.net
Thanks, And I appreciate the tip about the format to use for pic.

- Jun 12th 2007, 07:11 PMThePerfectHacker
- Jun 12th 2007, 07:13 PMwvmcanelly@cableone.net
Note taken,

Thanks