The problem goes something like this:
$$ f(x) = 4^x+8^x+9^x-2^x-6^x-12^x$$ Find the maximum value of f(x).
This question appeared on a test in which calculators were not allowed, so please refrain from using any graphing calculators or similar instruments.
I tried substituting $2^x=\alpha$ and $3^x=\beta$ to see if I could then observe something. It was of no help to me.
Any ideas or solutions would be appreciated.
On
You can compute the derivative: $$f'(x) = 4^x \ln 4 + 8^x \ln 8 + 9^x \ln 9 - 2^x \ln 2 - 6^x \ln 6 - 12^x \ln 12.$$ My instinct would be that a "nice" value of $x$ will probably make $f'(x) = 0$ (probably an integer). So, I would look for $x$ where this expression is $0$, by treating the $\ln 2$ and $\ln 3$ terms separately, hoping both go to $0$ at the same time. Factoring in this way, we get \begin{align} f'(x) &= (2 \cdot 4^x + 3\cdot 8^x - 2^x - 6^x - 2 \cdot 12^x) \ln 2 + (2 \cdot 9^x - 6^x - 12^x) \ln 3 \\ &= 2^x(2 \cdot 2^x + 3 \cdot 4^x - 1 - 3^x - 2 \cdot 6^x) \ln 2 + 3^x(2\cdot 3^x - 2^x - 4^x)\ln 3. \end{align} From eyeballing the $\ln 3$ term, $x = 1$ looks plausible, and we can confirm it works for the $\ln 2$ term as well. So, $x = 1$ is a stationary point.
Is $x = 1$ a global maximum? Yes. We can see this by showing the derivative is negative for $x > 1$ and positive for $x < 1$.
First, recall that the map $y \mapsto y^x$ is strictly convex if $x > 1$, and strictly concave if $x < 1$. In particular, for $x > 0$, we have $\frac{2^x + 4^x}{2} > 3^x$, hence $2 \cdot 3^x - 2^x - 4^x < 0$. For $x < 1$, the inequality is reversed. Similarly, when $x > 0$, we have $2 \cdot 2^x - 1 - 3^x < 0$, but when $x < 0$, the inequality is reversed. We also have: $$3 \cdot 4^x - 2 \cdot 6^x = 3 \cdot 4^x\left(1 - \left(\frac{3}{2}\right)^{x-1}\right)$$ which is also negative when $x > 1$, and positive when $x < 1$. Thus, putting it all together, we see $f'(x) < 0$ for $x > 1$ and $f'(x) > 0$ for $x < 1$. Thus, $f$ achieves a global maximum at $x = 1$. The value of this maximum is $f(1) = 1$.
We can proceed from your idea, by setting $a=2^x$ and $b=3^x.$ Then, consider the expression
$$S=a^2+a^3+b^2-1-a-ab-a^2b$$
which is a quadratic with respect to $b$ and the discriminant is $\Delta =(a+1)^2(a-2)^2$, therefore:
$\displaystyle{S=(b-a-1)(b-a^2+1)=(3^x-2^x-1)(3^x-4^x+1)=f(x)-1}$ or $\displaystyle{f(x)-1=8^x\left[ \left(\frac{3}{2} \right)^x-\left(\frac{1}{2} \right)^x-1\right]\left[ \left(\frac{3}{4} \right)^x+\left(\frac{1}{4} \right)^x-1\right]}$
Now $\displaystyle{ g (x)=\left(\frac{3}{2} \right)^x-\left(\frac{1}{2} \right)^x-1}$ is obviously increasing with unique root 1 and
$\displaystyle{h(x)=\left(\frac{3}{4} \right)^x+\left(\frac{1}{4} \right)^x-1}$ is decreasing with unique root 1.
Therefore $f(x)-1$ has only one root, for $x=1$, and it is negative if $x\neq 1$.