Originally posted by: Miggi
Consider the function below. (Round the answers to two decimal places. If you need to use -
or
, enter -INFINITY or INFINITY.)
f(x) = 2x tan(x)
-p/2 < x < p/2
(a) Find the interval where the function is increasing.
( 1, 2)
Find the interval where the function is decreasing.
( 3, 4)
(b) Find the local minimum value.
5
(c) Find the interval where the function is concave up.
( 6, 7)
a)
f(x) = 2 x tanx---------------------------------(1)
f ' (x) = 2 x sec^2 x + 2 tanx
f ' (x) = 2(x sec^2 x + tanx) ---------------(2)
f ' (x) = 0 at
x sec^2 x + tan x = 0
x sec^2 x = -tanx
x/cos^2 x = -sinx/cosx
x = -sinx cosx
x = -1/2 * 2 sinx cosx
x = -1/2 * sin(2x)
sin(2x) = -2x
This is true only at x = 0 but f ' ( x ) is not defined at this point.
f ' (x) is continuous at all points at which f (x) is defined.Therefore, in any interval which does not contain x = 0, f ' (x) should have the same sign everywhere as changing sign would mean that f ' (x) is 0 somewhere, which has been shown to be not possible.
Consider a point x = -pi/4 in the interval -pi/2 < x < 0
f ' (x) = 2 [(-pi/4) * sec^2(-pi/4) + tan(-pi/4)]
On calculating we get f ' (x) < 0
Therefore f (x) is decreasing in -pi/2 < x < 0
In the interval 0 < x < pi/2,
x > 0, sec^2 x > 0 and tan x > 0
Therefore f ' (x) = 2(x sec^2 x + tanx) > 0
Therefore f(x) is increasing in 0 < x < pi/2
Find the interval where the function is increasing.
(0, pi/2)
Find the interval where the function is decreasing.
(-pi/2, 0)
b) x = 0 is the only point at which f ' (x) changes sign from negative to positive. Therefore there is local minimum at x = 0
f( 0 ) = 2 * 0 * tan0 = 0
(b) Find the local minimum value.
0
c) From equation (2),
f ' (x) = 2(x sec^2 x + tanx)
f '' (x) = 2(x * 2secx * secx * tanx + sec^2 x + sec^2 x)
f '' (x) = 2(2 * x sec^2 x * tanx + 2 * sec^2 x)
f '' (x) = 4 sec^ x (x tanx + 1)
For concave up, f '' (x) > 0
4 sec^ x (x tanx + 1) > 0
xtanx + 1 > 0 (as 4 sec^2 x cannot be negative anywhere)
x tanx > - 1
In both -pi/2 < x < 0 and 0 < x < pi/2, both x and tanx are of same sign.
Therefore x tanx > 0
Therefore x tanx > -1
Thus the condition for concave up is satisfied.
(c) Find the interval where the function is concave up.
(-pi/2, 0) U (0, pi/2)
( 9,infinity)
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