Given that f(x) = 5+px - x2 = 9 - (x + q )2 where q is bigger than zero for all values of x.

Question

a) calculate the values of the  constants p and q.

b) state the maximum value of f(x) and the value of x at which it occurs.

c) find the range of values of x for each of the following.

1: 4x is larger than x2

Answer 1

(a) 5+px-x²=9-(x+q)²,

5+px-x²=9-x²-2qx-q²,

5+px=9-2qx-q²,

q²+2qx+px-4=0.

If this is to be true for all x then the x terms must cancel out:

2q+p=0, p=-2q, and q²-4=(q-2)(q+2)=0.

Since q>0 then q=2 and p=-4.

(b) f(x)=9-(x+2)².

When x=-2 f(x)=9 (maximum).

(c) 4x>x², x²-4x<0, x(x-4)<0.

To satisfy the inequality, x and x-4 must be of opposite signs. If x<0 then x-4 is also negative;

so if x>0, x-4<0, that is, x<4, therefore 0<x<4 (x is in the interval (0,4)).