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Find the minimum value of the function $f(x)=3^{x^2-4x}$ without using calculus.

I know how to do it by using calculus but I was wondering if this can be achieved by using the am-gm inequality by taking the natural log on both sides so $\ln(f(x))=(x^2-4x)\ln(3)$ ?

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    $\begingroup$ The minimum occurs when $x^2-4x$ is a minimum. So you just have to find the minimum of a quadratic. $\endgroup$
    – Hugh
    Oct 23 '16 at 21:54
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$$\;\;f(x) = 3^{x^2-4x} = 3^{(x^2-4x+4)-4} = \frac{1}{3^4} \cdot 3^{(x-2)^2} \ge \frac{1}{3^4} \cdot 3^0 = \frac{1}{3^4}$$

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  • $\begingroup$ Why inequality is this? $\endgroup$ Oct 23 '16 at 22:11
  • $\begingroup$ @dydxx Because $u = (x-2)^2 \ge 0$ and $u \mapsto 3^u$ is an increasing function. $\endgroup$
    – dxiv
    Oct 23 '16 at 22:13
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Well, let's just consider the function $f(x) = 3^x$ for a moment. It's clear that for every $x<y$ that $3^x<3^y$ ie the smaller $x$ is, the smaller $3^x$ is. So, to minimize the function you gave us, you would have to find when $x^2-4x$ has it's minimum. We can do that in the following way: If the quadratic is $ax^2 + bx + c$, the minimum is at $-b/2a$, so the minimum of $x^2-4x$ is at $x = 2$. Plugging $x = 2$ into $x^2 - 4x$, we find that the minimum of $x^2-4x$ is $-4$. Therefore, the minimum to your equation is $3^{-4}$. I hope that this helps!

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  • $\begingroup$ How do we get that the minimum is at -b/(2a) without using calculus? $\endgroup$
    – Hugh
    Oct 23 '16 at 22:14
  • $\begingroup$ @Hugh I honestly have no idea. I'm assuming that this is someone who hasn't been taught calculus, but has been taught that formula. $\endgroup$ Oct 23 '16 at 23:28
  • $\begingroup$ $x^2-4x=(x-2)^2-4$ could be easier to see the minimum. $\endgroup$
    – Macavity
    Oct 24 '16 at 4:14
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$e^{\mbox{crud}}$ will be smallest when crud is smallest. Here, crud $=x^2-4x$, whose graph is a parabola. It will be smallest at its vertex. One way to find the vertex is to note that $x^2-4x$ crosses the $x$-axis at $x=0$ and $x=4$, which means the $x$-coordinate of the vertex is at $x=2$. This makes the minimum $3^{-4}={1\over 81}$.

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The function $y\mapsto 3^y$ is a strictly increasing function, hence $3^{x^2-4x}$ attains its minimum when $x^2-4x$ is at its minimum. The problem is then reduced to finding the minimum of $x^2-4x$.

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