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I have an expression

$$\mathbf{\frac{1}{3} \; a^{2} - 4 \; a + \frac{3}{4}}$$

I need to find the smallest value of the expression when $a$ can be any value from interval $\mathbf{ \left(-∞; ∞ \right)}$.

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  • $\begingroup$ i can not see an equation $\endgroup$ Sep 24, 2016 at 17:17
  • $\begingroup$ @Dr.SonnhardGraubner That is odd. I can see it fine. $\endgroup$
    – 6005
    Sep 24, 2016 at 17:17
  • $\begingroup$ do you mean $$\frac{1}{3}a^2-4a+\frac{3}{4}=0$$? $\endgroup$ Sep 24, 2016 at 17:18
  • $\begingroup$ By that, @Dr.SonnhardGraubner means there is no equals sign. What you have is a function, and you wish to minimize this function. $\endgroup$ Sep 24, 2016 at 17:18

4 Answers 4

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Note that \begin{align} \frac{1}{3}a^2-4a+\frac{3}{4} &= \left(\frac{a}{\sqrt{3}}\right)^2 - 2\left(\frac{a}{\sqrt{3}}\right)(2\sqrt{3}) + (2\sqrt{3})^2-12+\frac{3}{4} \\ & = \left( \frac{a}{\sqrt{3}}-(2\sqrt{3}) \right)^2 -\frac{48}{4}+\frac{3}{4} \end{align} Then the minimum value of the expression $\frac{1}{3}a^2-4a+\frac{3}{4}$ is equal to $-\frac{48}{4}+\frac{3}{4}=-\frac{45}{4}$ and occurs when $\dfrac{a}{\sqrt{3}}-2\sqrt{3}=0$, i.e. $a=6$.

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in generally:

Let $f(x)=ax^2+bx+c$, where $a>0$. Then $$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\ge c-\frac{b^2}{4a}$$ And $$ax^2+bx+c=c-\frac{b^2}{4a}\Leftrightarrow x=-\frac b{2a}$$

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Answer using calculus The derivative with respect to $a$ is $$ \frac{2}{3} a - 4. $$ So if you solve $\frac{2}{3} a - 4 = 0$ and plug in the value for $a$ into the original expresion, you will get your answer.

Note, that it is a quadratic function that "opens up", so this point where the derivative is zero will in fact be a minimum (and not a maximum or some other kind of point.)

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if you mean a function $$f(a)=\frac{1}{3}a^2-4a+\frac{3}{4}$$ then you will get $$f'(a)=\frac{2}{3}a-4$$ and $$f''(a)=\frac{2}{3}$$ You must solve the equation $$\frac{2}{3}a-4=0$$

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