Question

2x^2+4x+1=0 explain how to get maxima and minima value​

Answer 1

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

Let assume that

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

can be rewritten as

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

can be further rewritten as

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

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

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

As we know that,

$\rm \: {(x + 1)}^{2} \geqslant 0$

$\rm\implies \:\rm \: 2{(x + 1)}^{2} \geqslant 0$

$\rm\implies \:\rm \: 2{(x + 1)}^{2} - 1 \geqslant 0 - 1$

$\rm\implies \:\rm \: f(x) \geqslant - 1$

$\rm\implies \:f(x) \:only \: have\:minimum \: value$

$\rm\implies \:Minimum \: value \: of \: f(x) \: = \: - \: 1$

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Alternative Method :- Using Calculus

Let assume that

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

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

$\rm \: f'(x) = 4x + 4$

For maxima or minima,

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

$\rm \: 4x + 4 = 0$

$\rm \: 4x = - 4$

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

As,

$\rm \: f'(x) = 4x + 4$

So,

$\rm \: f''(x) = 4$

$\rm\implies \:\rm \: f''(x) > 0$

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

and

Minimum value is f( - 1)

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

$\rm \: = \:2 - 4 + 1$

$\rm \: = \:3 - 4$

$\rm \: = \: - \: 1$

$\rm\implies \:Minimum \: value \: of \: f(x) \: = \: - \: 1$

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Basic Concept Used :-

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.

Answer 2

Answer

Given:-

The Equation:-

2x² × 4x + 1 = 0

Let:-

f(x) = 2x² + 4x + 1

Doing the first derivative of the given function f(x) as :-

f'(x) = d/dx (2x²+4x+1)

»» f'(x) = 4x + 4

Now finding the root of f'(x) by putting the received function equal to zero as:-

f'(x) = 4(x+1)

»» 4(x+1) = 0

»» x + 1 = 0/4

»» x + 1 = 0

Now, subtracting 1 from both the sides as:

»» x + 1 - 1 = -1

»» x = -1

Now, substituting the value of x in the given equation as:

f(x) = 2(-1)²+4(-1)+1

»» f(x) = 2×1 - 4+1

»» f(x) = 2-3

»» f(x) = -1

Therefore, the minima are -1 and there are no maxima.