Show that $h(e)<f(e)$ where e is the exponential
Solution 1:
Hoping no mistake, for $x=e$ and $a=\frac 12$, we have $$f(e)=\frac{\left(\frac{1}{2}+e\right) \left(\left(\frac{1}{2}+e\right)^2 \left(\frac{15}{2}+e\right)-4\right)}{\left(\frac{5}{2}+e\right)^3}-8^{\frac{1}{-1- 2 e}} e^{1+\frac{1}{1+2 e}}$$ $$h(e)=\frac{2^{1-\frac{3}{1+2 e}} e^{\frac{1}{1+2 e}-1} \left(1+e \left(-1+2 e \left(-3-2 e+9 \log ^2(2)+6 \log (2)\right)+12 \log (2)\right)\right)}{(1+2 e)^4}$$
Computed exactly $$f(e)=0.007813852236\qquad \text{and} \qquad h(e)=0.007821413387$$ $$f(e)-h(e)=-7.56115 \times 10^{-6}$$
Then $h(e) > f(e)$