Answer:
f'(x) > 0 on [tex](-\infty ,\frac{1}{e})[/tex] and f'(x)<0 on[tex](\frac{1}{e},\infty)[/tex]
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:
[tex]f'(x)>0[/tex]
To find its decreasing interval :
[tex]f'(x)<0[/tex]
2) Then let's find the critical point of this function:
[tex]f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0[/tex]
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''[tex]=\frac{1}{e}[/tex]≈0.37 for e≈2.72
[tex]x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37[/tex]
3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
