Find the Local Maxima and Minima x^3-9x

Problem

x3−9*x

Solution

1. Find the first derivative of the function ƒ(x)=x3−9*x to determine the slope of the tangent line.

(d(x3)−9*x)/d(x)=3*x2−9

2. Identify critical points by setting the first derivative equal to zero and solving for x

3*x2−9=0

3*x2=9

x2=3

x=±√(,3)

3. Find the second derivative to apply the second derivative test for concavity.

(d(3)*x2−9)/d(x)=6*x

4. Test the critical points using the second derivative ƒ(x)″=6*x

ƒ(√(,3))″=6√(,3)

6√(,3)>0

⇒x=√(,3)* is a local minimum.

ƒ″*(−√(,3))=−6√(,3)

−6√(,3)<0

⇒x=−√(,3)* is a local maximum.

5. Calculate the y-coordinates by substituting the critical points back into the original function ƒ(x)=x3−9*x

ƒ(√(,3))=(√(,3))3−9*(√(,3))=3√(,3)−9√(,3)=−6√(,3)

ƒ*(−√(,3))=(−√(,3))3−9*(−√(,3))=−3√(,3)+9√(,3)=6√(,3)

Final Answer

Local Maximum: *(−√(,3),6√(,3)), Local Minimum: *(√(,3),−6√(,3))

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