Answer to Question #146876 in Calculus for honey

Question

f(x)=sin^2(x/3)
defined on the interval [−8.72,1.76].
a.) f(x) is concave down on the region(s)
b.) A global minimum for this function occurs at
c.) A local maximum for this function which is not a global maximum occurs at
d.) The function is increasing on the region(s)

Answer 1

Let`s start to solve problem:

a) We need to find second derivative of the function in order to find the regions of concave down.

f ^{`</sup>(x)=(sin<sup>2</sup>(x/3))<sup>`} = 2*sin(x/3)*cos(x/3)/3=sin(2*x/3)/3 ---> this is the first derivative of given function.

f^{``</sup>(x)=(sin(2*x)/3)<sup>``}=2*cos(2*x/3)/9 --->this is the second derivative of given function.

When the second derivative of function is negative, then the function is concave downward:

f^``(x)<0, then:

2*cos(2*x/3)/9<0

cos(2*x/3)<0

$\pi$/2+2*$\pi$*k<2*x/3<3*$\pi$/2+2*$\pi$*k --> k is an integer number

3*$\pi$/4+3*$\pi$*k<x<9*$\pi$/4+3*$\pi$*k

In this case, if we look integer numbers to put instead of k, there is only -1 and it satisfies the interval of the function.

Let`s put -1 instead of k in order to get exact interval for being concave down:

3*$\pi$/4+3*$\pi$*(-1)<x<9*$\pi$/4+3*$\pi$*(-1)

-(9*$\pi$)/4<x<-(3*$\pi$)/4 or

-7.06<x<-2.35 this is the interval for being concave down!

b) We need to find first derivative of the function and it should equal to zero in order to get global minimum. But we already calculated first derivative of the given function and now we find global minimum:

f ^{`</sup>(x)=(sin<sup>2</sup>(x/3))<sup>`} = 2*sin(x/3)*cos(x/3)/3=sin(2*x/3)/3 and it should equal to zero. After then we reach: x=3*$\pi$*k/2. We put it into given interval:

-8.72<x<1.76

-2.77*$\pi$<3*$\pi$*k/2<0.56*$\pi$ after simplifying:

-1.84<k<0.373 --> this means that k can take -1 and 0 becouse of that it was integer number. Then we put it into x:

x₁=-(3*pi)/2 and x₂=0 if we put them into function:

f(x₁)=sin²(x/3)=1 and f(x₂)=sin²(x/3)=0.

It is clear that 0 is a global minimum of gven function!

c) We need to find first derivative of function and equal to zero. After then, we need to put the values of x into the given function and also we need to put boundry points of x into function in order to find local extremum points. But we already find and puted values of x and it is enough to put boundries into the function:

f(-8.72)=sin²(-8.72/3)=0.0025 and

f(1.76)=sin²(1.76/3)=0.000104.

In this case, local maximum is 0.0025.

d) First derivative of the function should be equal or greater than zero in order to find increasing interval. But we already found first derivative of the function and it is enough to find interval:

f ^{`</sup>(x)=(sin<sup>2</sup>(x/3))<sup>`} = 2*sin(x/3)*cos(x/3)/3=sin(2*x/3)/3,

sin(2*x/3)/3>=0,

0+2*$\pi$*k<x<$\pi$+2*$\pi$*k, k can get -1 and 0 then

-8.72<x<-4.71 and 0<x<1.76, this is also satisfied by boundries.

Therefore we can take it as increasing intervals.