Given f(x)=sin(x)-2cos(x) on the interval [0,2pi]. Determine where the function is concave up and concave down.
f'(x) = cosx + 2sinx
f''(x) = -sinx + 2cosx
points of inflection"
f''(x) = 0
-sinx + 2cosx = 0
sinx = 2cosx
tanx = 2
x = 63.4 degrees or x = 243.4 degrees
these last two are our important x values, since they are where the curve switches between concave up and concave down
lets's pick values in between
f''(x) = sin0 - 2cos0 = -2, so at 0 it is concave upwards
f''(90) = sin90 - 2cos9- = 1, so at 90 degrees it is concave down
f''(270) = sin270 - 2cos270 = -1 , so at 270 degrees it is concave upwards again.
f''(360) = sin360 - 2cos360 = -2 , sure enough concave upwards
so concave up: from 0 < x < 63.4
concave down : 63.4 < x < 243.4
concave up : 243.4 < x < 360
Just noticed you probably wanted your answer in radians, my calculator was set to degrees.
No big deal, just repeat my calculations with radian settings
To find where the function is concave up and concave down, we need to determine the sign of the second derivative. The second derivative test tells us that if the second derivative is positive, the function is concave up, and if the second derivative is negative, the function is concave down.
Let's start by finding the first derivative of f(x):
f(x) = sin(x) - 2cos(x)
f'(x) = cos(x) + 2sin(x)
Next, we find the second derivative by taking the derivative of f'(x):
f''(x) = -sin(x) + 2cos(x)
Now, to determine where the function is concave up and concave down, we need to find the points where the second derivative changes sign. We can set f''(x) = 0 and solve for x:
-sin(x) + 2cos(x) = 0
Divide both sides of the equation by cos(x):
-tan(x) + 2 = 0
tan(x) = 2
To find the solutions, we can use the inverse tangent function:
x = arctan(2)
Now we can analyze the signs of the second derivative in different intervals:
Interval [0, arctan(2)]:
For x in this interval, the second derivative f''(x) = -sin(x) + 2cos(x) is positive, since tan(x) is positive. Therefore, the function is concave up.
Interval (arctan(2), 2π]:
For x in this interval, the second derivative f''(x) = -sin(x) + 2cos(x) is negative, as tan(x) is negative. Hence, the function is concave down.
In summary, the function f(x) = sin(x) - 2cos(x) is concave up on the interval [0, arctan(2)], and concave down on the interval (arctan(2), 2π].