floymdiT

2021-06-04

Find the point at which the gradient of the curve with equation $y=3{x}^{2}-4x+1$ is zero.

sweererlirumeX

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

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