Question

Hola brainlians.


How do you find the maxima and minima of any function?

Let's say we are given a function,
f(x) = x³-6x²+9x+15.

How do you find it's maxima and minima? Can we find it's maxima and minima?

Can we find maxima and minima of any function? Yes/no, give reason for your choice.

Need detailed explanation.

​​

Answer 1

Step-by-step explanation:

Correct option is

C

15.4 m

∴v=v

x

i

^

+v

y

j

^

v=i×u

x

cosθ

i

^

+u

x

×cosθ×tan(θ/2)

j

^

v

2

=u

2

cos

2

θ+u

2

cos

2

θ×tan

2

θ/2

v

2

=u

2

cos

2

θ(1+tan

2

θ/2)

v

2

=u

2

cos

2

θ×sec

2

θ/2

∴ Radius of curvature =

gcosθ/2

u

2

cos

2

θ×sec

2

θ/2

=

4×3×10×

3

400×1×4×2

3

3

80

Answer 2

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

Basic Concept Used to find maxima or minima :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

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

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

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

Given function is

$\rm :\longmapsto\:f(x) = {x}^{3} - {6x}^{2} + 9x + 15$

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

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}( {x}^{3} - {6x}^{2} + 9x + 15 )$

We know,

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

So, using this, we get

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

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

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

$\rm :\longmapsto\:f'(x) =3[x(x - 3) - 1(x - 3)]$

$\rm :\longmapsto\:f'(x) =3[(x - 3) (x - 1)]$

Now, for maxima or minima

$\rm :\longmapsto\:f'(x) = 0$

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

$\bf\implies \:x = 1 \: \: or \: \: x = 3$

These are the points of Local maxima or Local minima.

To check out these points, we use the concept of double differentiation.

Now, we have

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

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

$\rm :\longmapsto\:\dfrac{d}{dx}f'(x) = \dfrac{d}{dx}[ {3x}^{2} - 12x + 9]$

$\rm :\longmapsto\:f''(x) = 3(2x - 3)$

Now,

$\rm :\longmapsto\:f''(1) = 3(2 - 3) = - 6 < 0$

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

and Maximum value is

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

$\bf\implies \:f(1) = 19$

Again,

$\rm :\longmapsto\:f''(3) = 3(6 - 3) = 9 > 0$

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

and Minimum value is

$\rm :\longmapsto\:f(3) = {3}^{3} - 6 {(3)}^{2} + 9(3) + 15$

$\rm :\longmapsto\:f(3) = 27 - 54 + 27 + 15$

$\bf\implies \:f(3) = 15$

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

The necessary condition for any function f(x) to have maxima or minima at any point x = a is f'(a) = 0.

For example :-

Consider the function

$\rm :\longmapsto\:f(x) = {e}^{x}$

So,

$\rm :\longmapsto\:f'(x) = {e}^{x} \: > 0 \: \forall \: x \: belongs \: to \: real \: number$

So,

$\rm :\longmapsto\:f(x) = {e}^{x} \: can \: neither \: have \: maxima \: nor \: minima$