Find the Local Maxima and Minima f(x)=2x^3-12x^2-72x+2017

Problem

ƒ(x)=2*x3−12*x2−72*x+2017

Solution

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

d(ƒ(x))/d(x)=6*x2−24*x−72

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

6*x2−24*x−72=0

3. Solve for x by factoring the quadratic equation.

6*(x2−4*x−12)=0

6*(x−6)*(x+2)=0

x=6,x=−2

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

d2(ƒ(x))/(d(x)2)=12*x−24

5. Test the critical point x=6 by substituting it into the second derivative.

ƒ(6)″=12*(6)−24=48

Since ƒ(6)″>0 the function is concave up, indicating a local minimum.

6. Test the critical point x=−2 by substituting it into the second derivative.

ƒ″*(−2)=12*(−2)−24=−48

Since ƒ″*(−2)<0 the function is concave down, indicating a local maximum.

7. Calculate the y-values for each critical point by substituting them back into the original function ƒ(x)

ƒ(6)=2*(6)3−12*(6)2−72*(6)+2017=1585

ƒ*(−2)=2*(−2)3−12*(−2)2−72*(−2)+2017=2097

Final Answer

Local Maxima: *(−2,2097), Local Minima: *(6,1585)

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