Question 579322
{{{ f(x)=3cos(x)}}} is continuous, but it does have maxima and minima.
To find where the function increases, decreases, or has a maximum or minimum, you need to look at its derivative.
To find curvature, points of inflection, concavity, you need to look at the second derivative.
The first derivative is
f'(x)={{{-3sin(x)}}}
It is zero at {{{x=0}}}, {{{x=pi}}}, {{{x=2pi}}}, etc
Those would be maxima or minima of f(x).
The derivative is positive when {{{-pi<x<0}}}, {{{pi<x<2pi}}}, {{{3pi<x<4pi}}}, {{{5pi<x<6pi}}} , etc. Those are the intervals where f(x) is increasing. Where {{{0<x<pi}}}, {{{2pi<x<3pi}}, {{{4pi<x<5pi}}}, the derivative is negative and the function decreases.
Putting it all together, the function increases from {{{-pi}}} to 0; has a maximum at x=0; decreases from there to {{{pi}}}; has a minimum at {{{x=pi}}}; increases from there to {{{2pi}}}; has a maximum at {{{x=2pi}}}, and so on. The value of the function is {{{f(x)=3}}} at all maxima and {{{f(x)=-3}}} at all minima.
The second derivative is f”(x)={{{-3cos(x)}}}.
It goes through zero and changes signs at {{{x=pi/2}}}, {{{x=3pi/2}}}, {{{x=5pi/2}}}, etc. Those are points of inflection of f(x). The value of the first derivative at those points is either 3 or -3, which gives you the slope of the tangent at those points.
The second derivative is negative between {{{-pi/2}}} and {{{pi/2}}}, positive between {{{pi/2}}} and {{{3pi/2}}}, negative between {{{3pi/2}}} and {{{5pi/2}}}, and so on. As a result the function is concave downwards between {{{-pi/2}}} and {{{pi/2}}}, concave upwards between {{{pi/2}}} and {{{3pi/2}}}, concave downwards between {{{3pi/2}}} and {{{5pi/2}}}, and so on.