# Math Help - Families of curves

1. ## Families of curves

Find a function of the form $y=ax^b lnx$ where a and b are nonzero constants and a local max at $(e^2, 6e^-1)$

So I'm not sure where to start on this one. I'm not too sure what all the different families of curves are, I found this website:Exponential and Logarithmic Function Gallery
but I can't tell which one of those my equation fits into.

Can I get some advice of where to start?

2. Hello,

First of all, you know that the point $(e^2,6e^{-1})$ belongs to the function.

So :

$\underbrace{y}_{6e^{-1}}=a \ \underbrace{x^b}_{e^{2b}} \ \ln(\underbrace{x}_{e^2})$

So $6e^{-1}=ae^{2b} \ln(e^2)=2ae^{2b}$

Here is a first equation involving a & b.

Then, as there is a maximum at this point, the derivative is null here. So find the derivative for y by x.

$\frac{dy}{dx}=abx^{b-1} \ln(x)+\frac{ax^b}{x}=abx^{b-1} \ln(x)+ax^{b-1}$ (used the product rule)

$\frac{dy}{dx}=ax^{b-1} [b+1]$

If $x=e^2$, $\frac{dy}{dx}=0$

So do the same as in the first part.

3. Here is what I got for an answer:

y = 3e^(-7) x^3 lnx

so a = 3e^-7 , b = 3

I took the equation you had got,

$
$
$\frac{dy}{dx}=ax^{b-1} [b+1]
" alt="\frac{dy}{dx}=ax^{b-1} [b+1]
" />

and plugged in e^2 for x, so

[tex] 0 = ae^2 [b+1]

so 2 = b-1, b=3
I then plugged that into the original equation, and got 3e^-7=a

but that equation I tried was wrong, so I'm not sure what I'm doing wrong. Any help?

4. Sorry for double post.

Can I get anymore help on this one? I'm still pretty lost.

5. Originally Posted by billabong7329
...
Ok, either I'm tired either what you wrote was not very clear... sorry about that, will try to explain..

Local max at $(e^2, 6e^{-1})$.
This means that this point belongs the curve.

So $f(e^2)=6e^{-1}$

$f(e^2)=a (e^2)^b \ln(e^2)=ae^{2b} \ln(e^2)$

Let's get rid of $\ln(e^2)$.
We know that $\ln(e^b)=b$
So $\ln(e^2)=2$

Hence $f(e^2)=6e^{-1}=2ae^{2b}$
Multiplying each side by $e^1$, and dividing by 2, we have :

$3*1=3=ae^{2b+1}$ (1)

Now, we know that there is a maximum at this point. This means that $f'(e^2)=0$

$f'(x)=(ax^b \ln(x))'=a(x^b \ln(x))'$
Here, use the product rule...

$f'(x)=a (bx^{b-1} \ln(x)+\frac{x^b}{x} )=a (bx^{b-1} \ln(x)+x^{b-1} )=a \ x^{b-1} (b \ln(x)+1)$

$f'(e^2)=a \ (e^2)^{b-1} (b \ln(e^2)+1)=a \ e^{2(b-1)} (2b+1)=0$

As the exponential function is never equal to 0 and a is a constant different from 0, $2b+1=0 \Longrightarrow b=\frac{-1}{2}$

Replacing in (1) :

$3=ae^{2b+1} \Longrightarrow 3=ae^0=a$

So $a=3$

This is only the way to do, it seems that I've made a mistake (dunno where...), according to your answer... So check my demonstration, this will make a good exercise, trust me