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 $f(x)$

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.
