Question

Find maximum/maximum value of y in the functions given below (a) y=5 - (x -1)^(2) (b) y =4x ^(2) - 4x + 7 (c) y= x^(3) - 3x y =x^(3) - 6x^(2) + 9x + 15 (e) y = (sin 2x - x), where - (pi)/(2) le xxle(pi)/(2)

Answer 1

The maximum/maximum value of y in the functions given below  are:

(a) y=5 - (x -1)^(2) is max (y) = 5

(b) y =4x ^(2) - 4x + 7 min(y) is 6

(c) y= x^(3) - 3x is  max(y) = +2 and min(y) = -2

(d)  y =x^(3) - 6x^(2) + 9x + 15 is min (y ) = 15 and max (y) =19

(e) y = (sin 2x - x) has no local maxima or minima

(a)  y=5 - (x -1)^(2) is 5

• ==> y' = - 2( x - 1 ) = 0
• x = 1
• y'' = -2 < 0 , therefore y will be local maximum at x = 1
• max (y) = 5 - (1 - 1 )^2 = 5

(b)  y = 4x ^(2) - 4x + 7 is 6

• ==> y' = 8x - 4 = 0
• x = 1/2
• y'' = 8 > 0
• Therefore it has only local minima
• min (y) = 4 (1/2)^2 - 4 (1/2) + 7 = 1 - 2 + 7 = 6

(c) y= x^(3) - 3x is

• ==> y' = 3x^2 - 3 = 0
• x = ±1
• y'' = 6x
• y'' = 6 when x =1 and -6 when x = -1
• Therefore local  minimum at x = 1 and local maximum at x = -1
• min(y) = 1 - 3 = -2
• max(y) = -1 + 3 = 2
• Cubic functions have global maxima and global minima at ±∞

(d) y =x^(3) - 6x^(2) + 9x + 15 is

• ==> y' = 3x^2 -12x + 9 = 0 ==>
• x^2 - 4x + 3 = 0 ==> x^2 - 3x - x +3 = x ( x -3 ) -1( x - 3 ) = (x -1 )( x -3) = 0
• x =1 , x =3
• y'' = 6x -12
• y'' = -6 , local maxima at x= 1 and local minima y'' = 6 , x =3
• min (y ) = 27 - 6x9 + 9x3 + 15 = 15
• max (y) = 1 - 6 + 9 + 15 = 19
• Cubic functions have global maxima and global minima at ±∞ .

(e) y = sin 2x - x

• ==> y' = cos2x - 1 = 0
• 2x = nπ
• x = nπ/2
• y'' = -sin2x = -sinnπ = 0
• Therefore no local maxima or minima. From graph we can infer that it actually has infinite local maxima or minima.