Solve 4x_1^2-6sqrt{3}x_{1}+3.<0

$$4 x _ { 1 } ^ { 2 } - 6 \sqrt { 3 } x _ { 1 } + 3 . < 0$$

Differentiate w.r.t. x_1

$$8x_{1}-6\sqrt{3}$$

Steps Using Derivative Rule for Sum

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$2\times 4x_{1}^{2-1}+\left(-6\sqrt{3}\right)x_{1}^{1-1}$$

Multiply 2 times 4.

$$8x_{1}^{2-1}+\left(-6\sqrt{3}\right)x_{1}^{1-1}$$

Subtract 1 from 2.

$$8x_{1}^{1}+\left(-6\sqrt{3}\right)x_{1}^{1-1}$$

Subtract 1 from 1.

$$8x_{1}^{1}+\left(-6\sqrt{3}\right)x_{1}^{0}$$

For any term t, t^{1}=t.

$$8x_{1}+\left(-6\sqrt{3}\right)x_{1}^{0}$$

For any term t except 0, t^{0}=1.

$$8x_{1}+\left(-6\sqrt{3}\right)\times 1$$

For any term t, t\times 1=t and 1t=t.

$$8x_{1}-6\sqrt{3}$$

Evaluate

$$4x_{1}^{2}-6\sqrt{3}x_{1}+3$$

Solution Steps

Multiply -1 and 6 to get -6.

$$4x_{1}^{2}-6\sqrt{3}x_{1}+3$$