Find the point at which the gradient of the curve with equation y=3x2−4x+1 is zero.

Given equation ${\displaystyle y=3x^2-4x+1}$
Gradient of the equation is $\frac{d}{dx}(3x^2-4x+1)$
Apply the Sum/Difference Rule: $(f+-g)=f+-g$' ${\displaystyle =\left(\frac{d}{dx}\right)\left(3x^2\right)-\left(\frac{d}{dx}\right)\left(4x\right)+\left(\frac{d}{dx}\right)\left(1\right)=6x-4}$
Given gradient is zero. Therefore, 6x-4=0 ${\displaystyle x=\frac{2}{3}}$
At x=0.66666... the gradient of the equation will be zero.