May 9th, 2014, 07:27 PM  #1 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus  Minima
What is the minimum value of the function $\displaystyle f(x) = 2ax^3  9ax^2 + 5a$ on the closed interval $\displaystyle [1, 2]$? State all of the possible minimums in terms of how $\displaystyle a$ varies.

May 9th, 2014, 08:06 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond 
Differentiate f(x) once. There is an extremum at x = 0 which is a minimum if a is negative. The minimum of the function on [1, 2] is then 5a.

May 9th, 2014, 08:13 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 
\begin{align*} f(x) &= 2ax^3  9ax^2 + 5a \\ f'(x) &= 6ax^2  18ax \\ \end{align*} So at a turning point: \begin{align*} 6ax^2 &= 18ax \\ x &= \begin{cases}3 &\text{or} \\ 0 \\ \end{cases} \\ \end{align*} So, when $a \gt 0$, we have a minimum at $x = 0$. When $a \lt 0$, we have only a local maximum in [1, 2], so the minimum value must be at one of the endpoints. $$f(1) = 2a 9a + 5a = 6a \\ f(2) = 16a  36a + 5a = 15a \\ f(1) \lt f(2)$$ since $a \lt 0$. 
May 9th, 2014, 08:42 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond  
May 9th, 2014, 09:15 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 
Oops! I got a sign error on the other root. Never mind, the method is clear enough.


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