Find range of $f(x)=3^x+5^x-8^x$ Find range of $f(x)=3^x+5^x-8^x$.
My attempt:
On observation one sees that $f(1)=0$.
On taking $g(x)=\left(\frac{3}{8}\right)^x+\left(\frac{5}{8}\right)^x-1$ and then observing that
$g'(x)=\left(\frac{3}{8}\right)^x \ln\left(\frac{3}{8}\right)+\left(\frac{5}{8}\right)^x \ln\left(\frac{5}{8}\right)<0$ 
we can conclude that $f(x)$ is negative for $x\in (1,\infty)$.
Also $\lim_{x\rightarrow-\infty}f(x)=0$.
Only thing to be done is to find maximum value of $f(x)$
 A: Considering the function $$f(x)=3^x+5^x-8^x$$ by inspection we have $$f(-\frac 12)=-\frac{1}{2 \sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\approx 0.67101$$ $$f(0)=1$$ $$f(\frac 12)=-2 \sqrt{2}+\sqrt{3}+\sqrt{5}\approx 1.13969$$ $$f(\frac 34)=-4 \sqrt[4]{2}+3^{3/4}+5^{3/4}\approx 0.86638$$ So, as already answered by Ahmed S. Attaalla, we can safely start Newton method for the zero of $f'(x)$ using $x_0=\frac 12$.
The iterates will then be
$$\left(
\begin{array}{cc}
n & x_n \\
 1 & 0.412626 \\
 2 & 0.397630 \\
 3 & 0.397236 \\
 4 & 0.397235
\end{array}
\right)$$ and, for the converged solution, $f(x)\approx 1.15811$ is the maximum.
From a numerical point of view, instead of looking for the zero of $$f'(x)=\ln 3 (3^x)+\ln 5 (5^x)-\ln 8 (8^x)$$ it would be more efficient to look at at zero of $$g(x)=x \log (8)-\log \left(\frac{3^x \log (3)+5^x \log (5)}{\log (8)}\right)$$ which is almost a straight line in the area of interest. Using the same $x_0=\frac 12$, the iterates would then be instead 
$$\left(
\begin{array}{cc}
n & x_n \\
 1 & 0.396751 \\
 2 & 0.397235 
\end{array}
\right)$$
As you can see, the first iterate is "almost" the solution.
A: $$f'(x)=\ln 3 (3^x)+\ln 5 (5^x)-\ln 8 (8^x)$$
Does not have a zero in term of elementary functions so you'll have to resort to numerical methods of approximation.
Newtons method for example,
$$x_{n+1}=x_{n}-\frac{\ln 3 (3^{x_n})+\ln 5 (5^{x_n})-\ln 8 (8^{x_n})}{(\ln 3)^2 3^{x_n} (\ln 5)^2 5^{x_n}-(\ln 8)^2 8^{x_n}}$$
Let $x_1$ be a number of your choice that lets the sequence converge.
Then by considering the signs of $f'(x)$ and the end behavior of $f(x)$ you may show:
$$\text{max} f(x)=f(\lim_{n \to \infty} x_n)$$
For the minimum consider:
$$\lim_{x \to \infty} (3^x+5^x-8^x)=\lim_{x \to \infty} (8^x((\frac{3}{8})^x+(\frac{5}{8})^x-1))$$
Which we consider because after proving $f$ to be decreasing on $x \in (\lim_{n \to \infty} x_n,\infty)$ or the easier $x \in (1,\infty)$, that leaves the question decreasing to some bound or decreasing to $-\infty$?
A: If you draw the function, it looks like this. So its range is $(f_{max}, -\infty)$

following is to analyze the function $f(x)=3^x+5^x-8^x$.


*

*$\lim\limits_{x\to -\infty} f(x) =0 $

*$\lim\limits_{x\to +\infty} f(x) =\lim\limits_{x\to +\infty} 8^x((\frac38)^x+(\frac 58)^x-1)=- \infty $

*$f(1) = 0$

*$f(0) = 1$


derivative of $g(x)=3^x+5^x-8^x$ is $g'(x)=ln3(3^x)+ln(5)(5^x)$. The derivative is positive and monotonically increasing. 
derivative of $h(x)=8^x$ is $h'(x)=ln8(x^8)$. The derivative is positive and monotonically decreasing.
So if there is any root, there is only one for $g'(x)=h'(x)$ or $f'(x)=0$. And this root is global optima. Because the minimum is $-\infty$, this point should be the global maximum. 
Because $f'(0)=\ln3+\ln5-\ln8=0.62$ and $f'(1)=3\ln3+5\ln5-8\ln8=-0.529$, the root of $f'(0)$ should be between $(0,1)$. Keep using bisection method, (or more advanced, using Newton method), you will find the maximum value at 0.397 with $f_{max}=1.158$.
Answer for range is $(-\infty, 1.158)$
