Question 1192776
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{{{f(x)=2cos(x)+3sin(x)}}}<br>
{{{df/dx=-2sin(x)+3cos(x)}}}<br>
{{{d^2f/dx^2=-2cos(x)-3sin(x)}}}<br>
The stationary point is where the derivative is zero.<br>
{{{-2sin(x)+3cos(x)=0}}}
{{{2sin(x)=3cos(x)}}}
{{{sin(x)/cos(x)=3/2}}}
{{{x = tan^(-1)(3/2)}}} ~ 0.983 radians<br>
There is little value in evaluating the function by hand when x has the value arctan(3/2); use a graphing calculator to find the approximate y value at the stationary point.<br>
The nature of the stationary point is determined by the sign of the second derivative.  The graphing calculator will show that the stationary point is a maximum.<br>
However, you can tell that the stationary point will be a maximum by looking at the second derivative of the function.  The stationary point is when x is between 0 and pi/2, so we are in the first quadrant, where sin(x) and cos(x) are both positive; so the sign of the second derivative is clearly negative, which means the stationary point is a maximum.<br>
A graph, for x from -pi/2 to 3pi/2, showing a maximum at approximately (1,3.6)....<br>
{{{graph(400,400,-pi/2,3pi/2,-5,5,2cos(x)+3sin(x))}}}<br>