Quadratic - Extreme Values

The quadratic equation of the form $$\mid x^2 - 4x - 13 \mid = r$$ has its minimum value at x = -b/2a, and hence does not vary irrespective of the value of x.

Hence at x = 2 the quadratic equation has its minimum.

Considering the quadratic part : $$\left|x^2-4\cdot x-13\right|$$. as per the given condition, this must-have 3 real roots.

The curve ABCDE represents the function $$\left|x^2-4\cdot x-13\right|$$. Because of the modulus function, the representation of the quadratic equation becomes :

ABC'DE. 

There must exist a value, r such that there must exactly be 3 roots for the function. If r = 0 there will only be 2 roots, similarly for other values there will either be 2 or 4 roots unless at the point C'.

The point C' is a reflection of C about the x-axis. r is the y coordinate of the point C' :

The point C which is the value of the function at x = 2, = $$2^2-8-13$$

= -17, the reflection about the x-axis is 17.

Alternatively,

$$\mid x^2 - 4x - 13 \mid = r$$ .

This can represented in two parts :

$$x^2-4x-13\ =\ r\ if\ r\ is\ positive.$$

$$x^2-4x-13\ =\ -r\ if\ r\ is\ negative.$$

Considering the first case : $$x^2-4x-13\ =r$$

The quadraticequation becomes : $$x^2-4x-13-r\ =\ 0$$

The discriminant for this function is : $$b^2-4ac\ =\ 16-\ \left(4\cdot\left(-13-r\right)\right)=68+4r$$

SInce r is positive the discriminant is always greater than 0 this must have two distinct roots.

For the second case :

$$x^2-4x-13+r\ =\ 0$$ the function inside the modulus is negaitve

The discriminant is $$16\ -\ \left(4\cdot\left(r-13\right)\right)\ =\ 68-4r$$

In order to have a total of 3 roots, the discriminant must be equal to zero for this quadratic equation to have a total of 3 roots.

Hence $$\ 68-4r\ =\ 0$$

r = 17, for r = 17 we can have exactly 3 roots.

$$f(x)=\frac{x^{2}+2x+4}{2x^{2}+4x+9}$$

If we closely observe the coefficients of the terms in the numerator and denominator, we see that the coefficients of the $$x^2$$ and x in the numerators are in ratios 1:2. This gives us a hint that we might need to adjust the numerator to decrease the number of variables.

$$f(x)=\frac{x^2+2x+4}{2x^2+4x+9}=\frac{x^2+2x+4.5-0.5}{2x^2+4x+9}$$

= $$\frac{x^2+2x+4.5}{2x^2+4x+9}-\frac{0.5}{2x^2+4x+9}$$

= $$\frac{1}{2}-\frac{0.5}{2x^2+4x+9}$$

Now, we only have terms of x in the denominator.

The maximum value of the expression is achieved when the quadratic expression $$2x^2+4x+9$$ achieves its highest value, that is infinity.

In that case, the second term becomes zero and the expression becomes 1/2. However, at infinity, there is always an open bracket ')'.

To obtain the minimum value, we need to find the minimum possible value of the quadratic expression.

The minimum value is obtained when 4x + 4 = 0 [d/dx = 0]

x=-1.

The expression comes as 7. 

The entire expression becomes 3/7.

Hence, $$[\frac{3}{7},\frac{1}{2})$$

First function $$f\left(x\right)=x^2-4cx+8$$

For this function $$a>0$$, so minimum value will occur at $$x=-\dfrac{b}{2a}=-\left(-\dfrac{4c}{2}\right)=2c$$

So, the minimum value of the function is = $$2c^2-4c\left(2c\right)+8c=-4c^2+8c$$

Second function $$g(x)=-x^{2}+3cx-2c$$

For this function $$a<0$$, so maximum value will occur at $$x=-\dfrac{b}{2a}=-\dfrac{\left(-3c\right)}{2}=\dfrac{3c}{2}$$

So, the maximum value of the function is = $$-\left(\dfrac{3c}{2}\right)^2+3c\left(\dfrac{3c}{2}\right)-2c=\dfrac{9c^2}{4}-2c$$

So, as per the given condition,

$$\dfrac{9c^2}{4}-2c<-4c^2+8c$$

or, $$\dfrac{9c^2}{4}+4c^2<8c+2c$$

or, $$\dfrac{25c^2}{4}<10c$$

or, $$\dfrac{5c^2}{4}<2c$$

or, $$5c^2<8c$$

or, $$5c^2-8c<0$$

or, $$c\left(c-\dfrac{8}{5}\right)<0$$

or, $$0<c<\dfrac{8}{5}$$

So, the value of $$c$$ which lies in this range is $$\dfrac{1}{2}$$