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Quadratic function f(x) - see details (more)

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f(x) and g(x) are both continuous functions (polynomials)

f(x) for x = 2, will be = 0

-22 + 2a + b = 0

2a + b = 4 ... (1)

also f(x) has only 1 root, hence discriminant = 0

a2 - 4*(-1)(b) = 0

a2 + 4b = 0

b =- a2 / 4

replacing value of b in (1)

2a - a2 / 4 = 0

8a - a2 = 0

a = 0 or a = 8

a = 0, b = 0

the function will be x2 = 0, a constant function (But question says that the function is quadratic)

hence a = 8

b = -16

f(x) = -x2 + 8x - 16 .... (2)

 

g(x) > 0 except for 2 <= x <= 3

thus g(x) has roots at x = 2 and x = 3

g(2) = 0

22 + 2c + d = 0

2c + d = -4 ... (3)

g(3) = 0

3c + d = -9 ... (4)

from 3 and 4

c = -5

d = 6

g(x) = x2 - 5x + 6 ... (5)

since h(x) is max [g(x), h(x)]

for h(x) we need to find the intersection of f(x) and g(x)

from 2 and 5

x2 - 5x + 6 = -x2 + 8x - 16

2x2 - 13x + 22 = 0

The solution is imaginary

the functions never intersect

g(x) is always > f(x)

h(x) = g(x)

value of x at which h(x) is minimum

g'(x) = 0

2x - 5 = 0

x = 5/2