Question

Q1) Find the maximum and minimum value of the function: x^3 - 3x^2 - 9x + 12​

Answer 1

Basic Concept Used :-

HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A FUNCTION

Let given function be f(x)

• Differentiate the given function f(x) w. r. t. x.

For maxima or minima,

• Put f'(x) = 0 and to get the critical points.

• Then find the derivative of f'(x) i.e. f''(x)

Apply these critical points in the second derivative.

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.

Let's solve the problem now!!

Given

$\rm :\longmapsto\:f(x) = {x}^{3} - {3x}^{2} - 9x + 12$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:f'(x) = 3 {x}^{2} - 6x - 9 - - - (1)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$

For maxima or minima,

• Put f'(x) = 0

$\rm :\longmapsto\: {3x}^{2} - 6x - 9 = 0$

$\rm :\longmapsto\: {x}^{2} - 2x - 3 = 0$

$\rm :\longmapsto\: {x}^{2} - 3x + x - 3 = 0$

$\rm :\longmapsto\:x(x - 3) + 1(x - 3) = 0$

$\rm :\longmapsto\:(x - 3)(x + 1) = 0$

$\rm :\implies\:x = 3 \: \: \: or \: \: \: x = - \: 1$

Now, Differentiate equation (1) w. r. t. x, we get

$\rm :\longmapsto\:f''(x) = 6x - 9$

Case :- 1

When x = 3, we get

$\rm :\longmapsto\:f''(3) = 6 \times 3 - 9 = 18 - 9 = 9$

$\rm :\longmapsto\:f''(3) > 0$

$\rm :\implies\:f(x) \: is \: minimum \: at \: x \: = \: 3$

and

$\rm :\longmapsto\:Minimum \: value \: = \: f(3)$

$\rm \: \: \: = \: \: {(3)}^{3} - 3{(3)}^{2} - 9 \times 3 + 12$

$\rm \: \: \: = \: \:27 - 27 - 27 + 12$

$\rm \: \: \: = \: \: - \: 15$

Hence,

$\: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: Minimum \: value \:of \: f(x) = \: - \: 15}$

Case :- 2

When x = - 1, we get

$\rm :\longmapsto\:f''( - 1) = 6 \times ( - 1) - 9 = - 6 - 9 = - 15$

$\rm :\longmapsto\:f''( - 1) < 0$

$\rm :\implies\:f(x) \: is \: maximum \: at \: x \: = \: - \: 1$

and

$\rm :\longmapsto\:Maximum \: value \: = \: f( - 1)$

$\rm \: \: \: = \: \: {( - 1)}^{3} - 3 {( - 1)}^{2} - 9( - 1) + 12$

$\rm \: \: \: = \: \: - 1 - 3 + 9 + 12$

$\rm \: \: \: = \: \:17$

Hence,

$\: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: Maximum \: value \:of \: f(x) = \: 17}$