Question

if y= 3x2-2x+1 find the value of x for which dy/dx=0​

Answer 1

y = 3x2-2x+1

$\frac{dy}{dx} = 2 \times 3 {x}^{2 - 1} - 1 \times 2 {x}^{1 - 1} + 0 \\ \\ \frac{dy}{dx} = 6x - 2 {x}^{0} + 0 \\ \\ \frac{dy}{dx} = 6x - 2$

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Answer 2

The given question is if y= 3x2-2x+1 find the value of x for which dy/dx=0

The given question is

$3{x}^{2} - 2x + 1 = y$

we have to find the value of x for which the value of the linear differentiation is zero(0)

This is the quadratic polynomial equation whose power is a square.

we have to find the value of x for which

$\frac{dy}{dx} = 0$

first, let us find the value of the linear differentiation.

the value will be

$\frac{dy}{dx} = \frac{d}{dx} (3 {x}^{2} - 2x + 1)$

The differentiation of the given equation is done by differentiating each term along with its coefficients.

$3( {2x}^{2 - 1} ) - 1 \times 2 {x}^{1 - 1} + 0 \\ {6x}^{1} - 2 {x}^{0} + 0$

The simplest expression is

$= 6x - 2$

It is given that

$\frac{dy}{dx} = 0 \\ so \: 6x - 2 = 0 \\ 6x = 2 \\ x = \frac{2}{6} \\ x = \frac{1}{3}$

Therefore the value of x is obtained as 1/3.

Hence, the value of x is obtained for the given polynomial.

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The rules used in the linear differentiation are given below

Linear differentiation:- [a.b(x) + c.d(y)]' = a.b'(x) + c.d'(x).

◙ Power rule:- (aˣ)' = x.aˣ⁻¹.

◙ Derivative of a variable is 1