Find the Local Maxima and Minima f(x)=x^3-3x^2+3

Problem

ƒ(x)=x3−3*x2+3

Solution

1. Find the first derivative of the function to identify the slope of the tangent line.

(d(x3)−3*x2+3)/d(x)=3*x2−6*x

2. Set the derivative to zero to find the critical points where the slope is horizontal.

3*x2−6*x=0

3. Factor the expression to solve for the variable x

3*x*(x−2)=0

4. Solve for x by setting each factor to zero.

x=0

x=2

5. Find the second derivative to apply the Second Derivative Test for concavity.

(d(3)*x2−6*x)/d(x)=6*x−6

6. Evaluate the second derivative at the critical point x=0

ƒ(0)″=6*(0)−6=−6

Since ƒ(0)″<0 the function is concave down, indicating a local maximum.

7. Evaluate the second derivative at the critical point x=2

ƒ(2)″=6*(2)−6=6

Since ƒ(2)″>0 the function is concave up, indicating a local minimum.

8. Calculate the y-coordinates by substituting the critical points back into the original function ƒ(x)

ƒ(0)=(0)3−3*(0)2+3=3

ƒ(2)=(2)3−3*(2)2+3=8−12+3=−1

Final Answer

Local Maximum: *(0,3), Local Minimum: *(2,−1)

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