Additional math (amath) differentiation of gradient, fins the equation of curve by finding the gradient and y intercept

Sun 12 noon Additional Math Tuition Class at Woodlands- Differentiation of gradient to find the equation of curve.

Any point on a curve that has stationary point (maximum/minimum) the gradient is equal to zero (dy/dx =0).

To determine if that particular point is maximum or minimum, do a second order differentiation d2y/dx2 , if d2y/dx2 <0 it is a maximum point and if  d2y/dx2 >0 it is a minimum point.

Integrate the expression dy/dx to find the equation of the curve , don’t forget to find the value of the constant (c) substitution the coordinates of the maximum point into the equation

Integrate \frac{{dy}}{{dx}}=3{{x}^{2}}-27

y=\int{{\left( {3{{x}^{2}}-27} \right)dx}}

y = x3 -27x + c

Find the stationary value of x, let dy/dx=0

\begin{array}{l}3{{x}^{2}}-27=0\\3{{x}^{2}}=27\\{{x}^{2}}=9\\x=\sqrt{9}\\x=3\ \text{or }x=-3\end{array}

Find the value of x at the maximum point,  d2y/dx2<0

\begin{array}{l}\text{Since }\frac{{dy}}{{dx}}=3{{x}^{2}}-27\\\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6x\end{array}

\begin{array}{l}\text{Sub x = 3}\\\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(3)=18\\\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}>0\text{ (Reject)}\end{array}

\begin{array}{l}\text{Sub x = -3}\\\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}=6(-3)=-18\\\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}<0\text{ }\end{array}

Find constant (c) by subsitution y=18 and x=-3

Since y = x3 -27x + c

Substitute y=18  and x=-3

18=(-3)-27(-3)+c

c=-36

Therefore the equation of the curve is y=x3 – 27x – 36

Additional Math (amath) differentiation of gradient to find the y-intercept and equation of curve.