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May 9th, 2014, 07:27 PM   #1
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Question 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.
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May 9th, 2014, 08:06 PM   #2
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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.
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May 9th, 2014, 08:13 PM   #3
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\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 $a \lt 0$.
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May 9th, 2014, 08:42 PM   #4
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Quote:
Originally Posted by v8archie View Post
So, when $a \gt 0$, we have a minimum at $x = 0$.
I disagree. Try graphing it.
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May 9th, 2014, 09:15 PM   #5
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Oops! I got a sign error on the other root. Never mind, the method is clear enough.
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