Question

Find the maximum minimum value of each of the following quadratic functions by completing the squares. In each state, the value of x at which the function is maximum minimum.
1) 3 - 4x - x². 2) 4x - 1 - x².​

Answer 1

$\large\underline{\sf{Solution-1}}$

Given quadratic function is

$\rm :\longmapsto\:f(x) = 3 - 4x - {x}^{2}$

can be rewritten as

$\rm :\longmapsto\:f(x) =- {x}^{2} - 4x + 3$

$\rm :\longmapsto\:f(x) =-({x}^{2} + 4x - 3)$

Now, On adding and Subtracting the square of half of the coefficient of x, we get

$\rm :\longmapsto\:f(x) =-({x}^{2} + 4x + {2}^{2} - {2}^{2} - 3)$

$\rm :\longmapsto\:f(x) =-[{(x + 2)}^{2} - 4-3]$

$\rm :\longmapsto\:f(x) =-[{(x + 2)}^{2} - 7]$

$\rm :\longmapsto\:f(x) =7-{(x + 2)}^{2}$

So, it implies f(x) assume its maximum value = 7, when x + 2 = 0

So, f(x) is maximum at x = - 2 and maximum value is 7.

$\large\underline{\sf{Solution-2}}$

Given quadratic function is

$\rm :\longmapsto\:f(x) = 4x - 1 - {x}^{2}$

can be rewritten as

$\rm :\longmapsto\:f(x) = - {x}^{2} + 4x - 1$

$\rm :\longmapsto\:f(x) = - ({x}^{2} - 4x + 1)$

Now, On adding and Subtracting the square of half the coefficient of x, we get

$\rm :\longmapsto\:f(x) = - ({x}^{2} - 4x + {2}^{2} - {2}^{2} + 1)$

$\rm :\longmapsto\:f(x) = - [({x}^{2} - 4x + 4) - 4 + 1]$

$\rm :\longmapsto\:f(x) = - [({x - 2)}^{2} - 3]$

$\rm :\longmapsto\:f(x) =3 - ({x - 2)}^{2}$

So, it implies f(x) assume its maximum value = 3, when x - 2 = 0

So, f(x) is maximum at x = 2 and maximum value is 3

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Additional Information

1. The quadratic polynomial f(x) = ax² + bx + c > 0, if a > 0 and b² - 4ac < 0, for every x is a real number.

2. The quadratic polynomial f(x) = ax² + bx + c < 0, if a < 0 and b² - 4ac < 0, for every x is a real number.

3. The quadratic polynomial f(x) = ax² + bx + c always have maximum value if a < 0 and there is no minimum value.

4. The quadratic polynomial f(x) = ax² + bx + c, always have minimum value if a > 0 and there is no maximum value.