Part A:
Given the function [tex]f(x)=x+2\cos(x)[/tex], the absolute maximum or minimum occurs when [tex]f'(x)=0[/tex].
[tex]f'(x)=0 \\ \\ \Rightarrow1-2\sin{x}=0 \\ \\ \Rightarrow2\sin{x}=1 \\ \\ \Rightarrow\sin{x}= \frac{1}{2} \\ \\ \Rightarrow x=\sin^{-1}{\frac{1}{2}}= \frac{\pi}{6} [/tex]
Using the second derivative test,
[tex]f''(x)=-2cosx \\ \\ \Rightarrow f''\left( \frac{\pi}{6} \right)=-2\cos{\left( \frac{\pi}{6} \right)}=-1.732[/tex]
Since the second derivative gives a negative number, the given function has a maximum point at [tex]x=\frac{\pi}{6}[/tex].
And the maximum point is given by:
[tex]f\left( \frac{\pi}{6} \right)=\frac{\pi}{6}+2\cos\left( \frac{\pi}{6} \right) \\ \\ =0.5236+2(0.8660)=0.5236+1.732 \\ \\ =\bold{2.256}[/tex]
i.e. [tex]\left(\frac{\pi}{6},\ 2.256\right)[/tex]
Part B:
Given the function [tex]f(x)=e^{-x}-e^{-2x}[/tex], the absolute maximum or minimum occurs when [tex]f'(x)=0[/tex].
[tex]f'(x)=0 \\ \\ \Rightarrow-e^{-x}+2e^{-2x}=0 \\ \\ \Rightarrow2e^{-2x}=e^{-x} \\ \\ \Rightarrow2e^{-x}=1 \\ \\ \Rightarrow e^{-x}=\frac{1}{2} \\ \\ \Rightarrow-x=\ln \frac{1}{2}=-0.6931 \\ \\ \Rightarrow x=0.6931[/tex]
Using the second derivative test,
[tex]f''(x)=e^{-x}-4e^{-2x} \\ \\ \Rightarrow f''(0.6931)=e^{-0.6931}-4e^{-2(0.6931)} \\ \\ =0.5-4e^{-1.386}=0.5-4(0.25)=0.5-1 \\ \\ =-0.5[/tex]
Since the second derivative gives a negative number, the given function has a maximum point at [tex]x=0.6931[/tex].
And the maximum point is given by:
[tex]f(0.6931)=e^{-0.6931}-e^{-2(0.6931)} \\ \\ =0.5-e^{-1.386}=0.5-0.25=\bold{0.25}[/tex]
i.e. (0.693, 0.25)
