find absolute extrema

Question

a) f(x)= (x^2)/(x^2+3), -1 less than or equal to x less than or equal to 1

b)h(x)=csc x, -pi/6 less than x less than or equal to pi/3

Answer 1

a) The rational function

The absolute extrema are where the first derivative equals zero.If it is a minimum the second derivative will be positive.If it is a maximum the second derivative will be positive.

So the extrema occur at x = 0 on the interval [- 1 ≤ x ≤ 1]

To determine if it is maximum or minimum ,we find the sign of the second derivative at x = 0.

Since it is positive, then x = 0 is a minimum.

Absolute minimum at (0,0).

Answer 2

b) h(x) = cscx

Apply derivative on each side with respect of x.

h'(x) = - cscx cotx

The absolute extrema are where the first derivative equals zero.If it is a minimum the second derivative will be positive.If it is a maximum the second derivative will be positive.

- cscx cotx = 0

- (1/sinx)(cosx/sinx) = 0

(cosx)/(- sin²  x) = 0

cos(x) = 0

cos(x) = cos(π/2)

Principal value x = π/2.

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer.

If n = 0 then x = 2(0)π ± π/2 = π/2, -π/2

If n = 1 then x = 2(1)π ± π/2 = 3π/2, 5π/2

If n = 2 then x = 2(2)π ± π/2 = 4π ± π/2= 9π/2, 7π/2

There is no solutions on the interval [-π/6, π/3].

Absolute extrema does not exist in the interval -π/6 ≤ x ≤ π/3.