# Stationary points - Solution

We select any continuous function e.g. t^3-3t^2+2*t-1 which we have plotted below, with -2<=t<4, using plotXpose.

## t^3-3t^2+2*t-1

Differentiating z= t^3-3t^2+2*t-1 (using the rules of differentiation) we get :

dz/dt= 3*t^2-6*t+2

and we can now plot that derivative function using plotXpose

## 3*t^2-6*t+2

Using Mode -> Zero finding and changing the accuracy to 10 significant figures we tap near each of the two zeros of this derivative function.

Solving 3*t^2-6*t+2=0: Success. The Newton-Raphson method has converged to the value
(0.4226497308,0.0000000000E+000).
The sequence of values found was
( n, t , y(t))
(0, 0.15686275065, 1.13264126373E+000)
(1, 0.38075695542, 1.50385844788E-001)
(2, 0.42123267442, 4.91485148547E-003)
(3, 0.42264799605, 6.00940594353E-006)
(4, 0.42264973081, 9.02788954704E-012)
(5, 0.42264973081, 4.44089209850E-016)
(6, 0.42264973081, 0.00000000000E+000)

At this point, where t is approx 0.4226497308, the plotted derivative is going from positive to negative therefore represents a maximum of the original function.

Solving 3*t^2-6*t+2=0: Success. The Newton-Raphson method has converged to the value
(1.577350269,1.7763568394E-015).
The sequence of values found was
( n, t , y(t))
(0, 1.7058823109, 4.94809510386E-001)
(1, 1.5890522806, 4.09477679041E-002)
(2, 1.5774665043, 4.02690641170E-004)
(3, 1.5773502809, 4.05236200152E-008)
(4, 1.5773502692, -1.77635683940E-015)
(5, 1.5773502692, 8.88178419700E-016)
(6, 1.5773502692, 1.77635683940E-015)

At this point, where t is approx 1.577350269, the plotted derivative is going from negative to positive therefore represents a minimum of the original function.

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