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 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: 7,968 Thanks: 1152 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: 7,691 Thanks: 2670 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 $a \lt 0$. Thanks from Mr Davis 97
May 9th, 2014, 08:42 PM   #4
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Quote:
 Originally Posted by v8archie So, when $a \gt 0$, we have a minimum at $x = 0$.
I disagree. Try graphing it.

 May 9th, 2014, 09:15 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,691 Thanks: 2670 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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