Question #830d3

1 Answer

The information on the increase and decrease intervals of a differentiable function are given by the first derivative, while the information about its concavity direction are given by the second derivative (if the function is twice differentiable).

In this case f:[0,2 pi] to mathbb{R} such that f(x)=7 sin x +7 cos x. This function is a composition of twice differentiable functions, so it's twice differentiable.

The first derivative is f'(x)=7 cos x - 7 sin x.
Let's find the values for which f'(x)>0:
7 cos x - 7 sin x >0
cos x >sin x
Let's look at the graphs of the two functions (cos x is green, sin x is red)

In [0,2 pi] they intersect when cos x = sin x hArr tan x=1 hArr x=pi /4 + k pi where k=0,1. So cos x >sin x hArr x in (0,pi/4) cup (5/4 pi, 2 pi) and cos x < sin x hArr x in (pi/4,5/4 pi).
By the increase/decrease test and the first derivative test :

  • f increases on (0,pi/4) cup (5/4 pi, 2 pi);
  • f decreases on (pi/4,5/4 pi);
  • the local maxima of f are at x=pi/4 and x=2 pi; their values are respectively f(pi/4)=7 sqrt(2) and f(2pi)=7;
  • the local minima of f are at x=0 and x=5/4 pi; their values are respectively f(0)=7 and f(5/4 pi)=-7 sqrt(2);

The second derivative is f''(x)=-7 sin x-7cos x.
Let's find the values for which f''(x)>0:
-7 sin x - 7 cos x >0
cos x < - sin x
As before, we can plot the two functions (cos x is green, -sin x is red)

In [0,2 pi] they intersect when cos x = - sin x hArr tan x=-1 hArr x=3 /4 pi + k pi where k=0,1. So cos x < -sin x hArr x in (3/4 pi, 7/4 pi) and cos x > -sin x hArr x in (0,3/4 pi) cup (7/4 pi, 2 pi).
By the concavity theorem:

  • f is concave upward on (3/4 pi, 7/4 pi);
  • f is concave downward on (0,3/4 pi) cup (7/4 pi,2pi);

The graph of this function is plotted in the following picture (warning: plot realized using different scaling factors on the two cartesian axes)