91影视

We have been given the function,

$f\left(x\right)=2^x(2^x鈭�1)$

We have to find the derivative of the given function

$f\left(x\right)=2^x(2^x鈭�1)$

Therefore,

$\begin{array}{rl}{f}^{\mathrm{鈥�}}\left(x\right)& =\frac{d}{dx}(2^{2x}鈭�2^x)\\ & =2^{2x}\ln 鈦�2脳2鈭�2^x\ln 鈦�2\\ & =2^{2x+1}\ln 鈦�2鈭�2^x\ln 鈦�2\end{array}$

Since the derivative is always defined and continuous, the critical points of the function are just the places where $f\text{'}\left(x\right)=0$; that is,

$\begin{array}{r}{2}^{2x+1}\ln 鈦�2鈭�2^x\ln 鈦�2=0\\ 2^x鈰�2^{x+1}\ln 鈦�2鈭�2^x\ln 鈦�2=0\\ 2^x\ln 鈦�2(2^{x+1}鈭�1)=0\\ 2^{x+1}鈭�1=0\end{array}$

$\begin{array}{rl}{2}^{x+1}& =1\\ 2^{x+1}& =2^0\\ x+1& =0\\ x& =鈭�1\end{array}$

These critical points divide the real number line into two parts$(鈭�\mathrm{鈭�},鈭�1),(鈭�1,\mathrm{鈭�})$

Here, for testing take the sign of $f\text{'}\left(x\right)$at one point in that interval.

Let the value of $x$be$x=-2,0$

$\begin{array}{rl}{f}^{\mathrm{鈥�}}(鈭�2)& =2^{2(鈭�2)+1}\ln 鈦�2鈭�2^{(鈭�2)}\ln 鈦�2\\ & =2^{鈭�4+1}\ln 鈦�2鈭�\frac{\ln 鈦�2}{4}\\ & =\frac{\ln 鈦�2}{8}鈭�\frac{\ln 鈦�2}{4}\\ & =鈭�\frac{\ln 鈦�2}{8}<0\\ f^{\mathrm{鈥�}}\left(0\right)& =2^{2\left(0\right)+1}\ln 鈦�2鈭�2^{\left(0\right)}\ln 鈦�2\\ & =2\ln 鈦�2鈭�\ln 鈦�2\\ & =\ln 鈦�2>0\end{array}$

Let us draw the sign chart,

The $f\text{'}$changes from negative to positive at $x=-1$, that is $f$changes from decreasing to increasing at $x=-1$. The function is concave up at $(-鈭�,鈭�)$.