finding values from power functions

**jojoferni244** · October 6th 2008, 07:23 PM

i have no clue how to do this problem. i need some help.

Given a power function of the form f(x)=ax^n, with f ' (3)=18 and f ' (6)=72, find n and a .

**Chris L T521** · October 6th 2008, 07:31 PM

Originally Posted by jojoferni244

i have no clue how to do this problem. i need some help.

Given a power function of the form f(x)=ax^n, with f ' (3)=18 and f ' (6)=72, find n and a .

Find f'(x):

$f'(x)=an x^{n-1}$

Now apply both conditions to this to come up with a system of equations:

$18=3^{n-1}\cdot a\cdot n$

$72=6^{n-1}\cdot a\cdot n$

Can you continue?

--Chris

**jojoferni244** · October 6th 2008, 07:34 PM

actually im stuck...can you help me please

**Chris L T521** · October 7th 2008, 12:52 PM

Originally Posted by Chris L T521

Find f'(x):

$f'(x)=an x^{n-1}$

Now apply both conditions to this to come up with a system of equations:

$18=3^{n-1}\cdot a\cdot n$

$72=6^{n-1}\cdot a\cdot n$

Can you continue?

--Chris

Originally Posted by jojoferni244

actually im stuck...can you help me please

Note that 4*18=72

So we can see that the last equation is $18=\tfrac{1}{4}\cdot6^{n-1}an$

Thus, $\tfrac{1}{4}6^{n-1}an=3^{n-1}an\implies \tfrac{1}{4}6^{n-1}=3^{n-1}\implies \tfrac{1}{4}=\frac{3^{n-1}}{6^{n-1}}\implies \tfrac{1}{4}=\left(\tfrac{3}{6}\right)^{n-1}$ $\implies \tfrac{1}{4}=\left(\tfrac{1}{2}\right)^{n-1}\implies \tfrac{1}{4}=\left(\tfrac{1}{2}\right)^{-1}\left(\tfrac{1}{2}\right)^n\implies \tfrac{1}{8}=\left(\tfrac{1}{2}\right)^n$

Can you take it from here? You will find an integer value for n.

Then plug this into either equation and solve for a.

--Chris

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