we have the function
[tex]f\left(x\right)=2^{x}\cos x[/tex]
Find out the critical points
so
Find out the first derivative
[tex]f^{\prime}(x)=ln2*2^xcosx-2^xsinx[/tex]
Equate the first derivative to zero
[tex]\begin{gathered} ln2*2^xcosx-2^xsinx=0 \\ ln2=\frac{2^xsinx}{2^xcosx} \\ \\ ln2=tanx \end{gathered}[/tex]
the value of the tangent is positive
that means
the angle x lies on the I quadrant or III quadrant
but remember that the interval is [0, pi]
therefore
The angle x lies on the quadrant
[tex]\begin{gathered} tanx=ln2 \\ x=0.2\pi\text{ radians} \end{gathered}[/tex]
The critical point is x=0.2pi radians
Find out the second derivative
[tex]f^{\prime}^{\prime}(x)=ln^22*2^xcosx-ln2*2^{x+1}*sinx-2^xcosx[/tex]
Evaluate the second derivative at x=0.2pi radians
The value of the second derivative is negative
so
The concavity is down
that means
The critical point is a local maximum in the given interval
