Quotient Rule: max of f(x) = (6x^2)/(5 + x^2) on [0, 3]

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Apply the quotient rule to find the maximum value of the function

\(\displaystyle \L f(x) = \frac{{6x^2 }}{{5 + x^2 }}\)

on the interval [0,3].

Please can you help? Thanks
 
Sure! What's the Quotient Rule? Can you apply it to find the derivative? That would be a good start.

One word of caution, with a restricted Domain, you MUST check the endpoints. There derivative doesn't exist there and it doesn't know if critical points exist.
 
Judging by your other thread, looks like you need some revision on derivatives.

If you don't know your quotient rules and your product rules (read up on it!):

\(\displaystyle \L\ \frac{6x^2}{5 + x^2}\ = 6 - \frac{30}{x^2 + 5}\\) for all real x, if that helps.
 
Here goes:

\(\displaystyle u = 6x^2 ,u' = 12x,v = 5 + x^3 ,v' = 3x^2\)

\(\displaystyle \frac{{dy}}{{dx}} = \frac{{12x(5 + x^3 ) - 3x^2 (6x^2 )}}{{(5 + x^3 )^2 }} = \frac{{60x + 18x^4 }}{{25 + 10x^3 + x^6 }}\)

I think this is the first derivative. But how do I find the maximum value? Do I substitute the values 0 and 3 into the original equation, or into the first derivative? Obviously, f(0) and f'(0) are going to 0, or am I missing something?

Thank you for all your help so far.
 
Your function seems to have changed from its original form. Just what is that exponent in the denominator?

You find where the first derivative is zero. Set the first derivative equal to zero and solve for x. It will only be a coincidence that x = 0. Do NOT substitute values for x. The point is to FIND the right values for x by solving using all that algebra you brought with you when you came to the calculus.
 
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