Change of variables when minimizing a function

[issacnewton]

Homework Statement

I am trying to minimize the function $f(a) = (1+4a^2)^3 \left( \frac{1}{4a^2} \right)^2$. Here we are given that $a>0$

Homework Equations

Definition of a minimum of a function

The Attempt at a Solution

Now the derivative here will be ugly and equating it to zero and solving it will be messy. So I did a substitution, $\alpha = 4a^2$. With this the function becomes $f(\alpha) = \frac{(1+\alpha)^3}{\alpha^2}$. The derivative is easier to calculate $$f'(\alpha) = \frac{(1+\alpha)^2}{\alpha^2}\left[ 3 - \frac{2(1+\alpha)}{\alpha} \right] $$ Now since $a>0$, we have $\alpha >0$ and when we equate $f'(\alpha)$ to zero, the equation just becomes $$\left[ 3 - \frac{2(1+\alpha)}{\alpha} \right] = 0$$ which is much easier to solve. $\alpha = 2$. And using the relationship between $a$ and $\alpha$, we can see that the function has a critical point when $a = \frac{1}{\sqrt{2}}$. With either first derivative test or the second derivative test, we can confirm that this is where the function has a local minimum. Now is it ok to do change of variable like I have done here in maxima minima problems. How do I justify this ?

Thanks
$\ddot\smile$

 

[fresh_42]

You have $f(a)=g(\alpha)=g(\alpha(a))$ and want to know, where $f'(a)=\frac{d}{da}f(a)=\frac{d}{da}g(\alpha(a))=0$. What does the chain rule tell you?

 

[issacnewton]

Chain Rule tells us that $\alpha'(a) g'(\alpha) = 0$

 

[fresh_42]

Chain Rule tells us that $\alpha'(a) g'(\alpha) = 0$

Yes. So the question is whether $\alpha'(a)$ can be zero or not. If not, then $f'=0$ and $g'=0$ are equivalent.

 

[issacnewton]

Here $\alpha'(a) = 8a \ne 0$ , which means that $g'(\alpha) = 0$ and we will solve for $\alpha$ and this will give us the critical value for $f$. I think this is clear now.