Find the Local Maxima and Minima f(x)=2x^3-3x^2-12x

Problem

ƒ(x)=2*x3−3*x2−12*x

Solution

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

d(ƒ(x))/d(x)=6*x2−6*x−12

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

6*x2−6*x−12=0

3. Solve for x by factoring the quadratic equation.

6*(x2−x−2)=0

6*(x−2)*(x+1)=0

x=2,x=−1

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

d2(ƒ(x))/(d(x)2)=12*x−6

5. Evaluate the second derivative at the critical points.

ƒ(2)″=12*(2)−6=18

ƒ″*(−1)=12*(−1)−6=−18

6. Determine the nature of the points. Since ƒ(2)″>0 there is a local minimum at x=2 Since ƒ″*(−1)<0 there is a local maximum at x=−1

7. Calculate the y-coordinates by substituting the x-values back into the original function ƒ(x)

ƒ(2)=2*(2)3−3*(2)2−12*(2)=16−12−24=−20

ƒ*(−1)=2*(−1)3−3*(−1)2−12*(−1)=−2−3+12=7

Final Answer

Local Maxima: *(−1,7), Local Minima: *(2,−20)

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