Find the absolute maximum and absolute minimum values of f on the given interval.
a) f(t)= t sqrt(36-t^2) [-1,6]
absolute max=
absolute min=
b) f(t)= 2 cos t + sin 2t [0,pi/2]
absolute max=
absolute min=

absolute max= (4.243,18)

absolute min =(-1,-5.916)

absolute max=(pi/6, 2.598)

absolute min = (pi/2,0)

Step-by-step explanation:

a)

To find max and minima in the given interval let us take log and differentiate

It is sufficient to find max or min of Y

In the given interval only 4.243 lies

And we find this is maximum hence maximum at  (4.243,18)

Minimum value is only when x = -1 i.e. -5.916

b)

Equate I derivative to 0

-2sint +1-2sin^2 t=0

sint = 1/2 only satisfies I quadrant.

So when t = pi/6 we have maximum

Minimum is absolute mini in the interval i.e. (pi/2,0)

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.