3. Use the second derivative test to find the local maxima and minina of the fol

Question

3. Use the second derivative test to find the local maxima and minina of the following functions. Also, determine the concavity and points of inflection.

(i) y = x^3 + 3x^2 2 ; (ii) y = (x^2/2) ln(x) ; (iii) y = x(x + 1)^1/3

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4. Show that arctan(x)+arctan(1/x) = /2 , (x not= 0) (Hint: Prove that d dx[arctan(x)+ arctan(1/x)] = 0).

5. The Intermediate Value Theorem (I.V.T.) shows that the functions (i) f(x) =

X^3 + x + 2 and (ii) g(x) = arctan(x) 1 have at least one root (i.e., c such that f(c) = 0 - check this). How can you show that this value of c is unique?

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Explanation / Answer

3. Use the second derivative test to find the local maxima and minina of the following functions. Also, determine the concavity and points of inflection.

(i) y = x^3 + 3x^2 2

dy/dx=-3x2+6x =0

==>-3x(x-2)=0 ==>x=0,x=2

d2y/dx2=-6x+6

d2y/dx2 at x=0 ==>-0+6 =6 >0 ==>local minima at x=0==>y=-0+0-2==>(x,y)=(2,-2)

d2y/dx2 at x=2 ==>-12+6 =-6 <0 ==>local maxima at x=2 ==>y=-2^3+3*2^22=2==>(x,y)=(2,2)

d2y/dx2=-6x+6 =0==>x=1==> inflection point at x=1. y=-1+3-2=0 ==>(x,y)=(1,0)

concave upwards==>d2y/dx2=-6x+6>0==>x>1==>x=(1,infinity)

concave downwards==>d2y/dx2=-6x+6<0==>x<1==>x=(-infinity,1)

(ii) y = (x^2/2) ln(x). domain is (0,infinity)

dy/dx=x- (1/x)

==>(x^2 -1)/x=0 ==>x=-1,x=1 but x=-1 not in domain

d2y/dx2=1- (-1/x2)

d2y/dx2=(x2+1)/x2

d2y/dx2 at x=1 ==>((1)2+1)/(1)2=2 >0 ==>local minima at x=1==>y= (1^2/2) ln(1) =1/2==>(x,y)=(1,1/2)

d2y/dx2=(x2+1)/x2 =0==>nevert happens==> inflection point doesnot exist

concave upwards==>d2y/dx2=(x2+1)/x2>0==>==>x=(0,infinity)

concave downwards==>d2y/dx2=(x2+1)/x2 <0==>neverhappens

(iii) y = x(x + 1)1/3 domain(-1,infinity)

dy/dx=(x + 1)1/3 +(x/3)(x + 1)-2/3

dy/dx=[3(x + 1) +x]/(3(x + 1)2/3)

dy/dx=[4x + 3]/(3(x + 1)2/3)

==>[4x + 3]/(3(x + 1)2/3)==>x=-3/4

d2y/dx2=(2(2x+3))/(9(x+1)(5/3))

d2y/dx2 at x=(-3/4) ==>(2(2(-3/4) +3))/(9((-3/4) +1)(5/3)) ==>2(3/2 )/(9(1/4)5/3) >0 ==>local minima at x=(-3/4)

==>y = (-3/4) ((-3/4) + 1)1/3==>y =-3(1/4)4/3

(x,y)=(-3/4-3(1/4)4/3)

concave upwards==>d2y/dx2=(2(2x+3))/(9(x+1)(5/3))>0==>x=(-1,infinity)

concave downwards==>d2y/dx2=(2(2x+3))/(9(x+1)(5/3))/x2 <0==>neverhappens

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