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Find the Inflection Points f(x)=2x^3+3x^2-180x

Problem

ƒ(x)=2*x3+3*x2−180*x

Solution

  1. Find the first derivative by applying the power rule to each term of the function.

d(ƒ(x))/d(x)=6*x2+6*x−180

  1. Find the second derivative by differentiating the first derivative with respect to x

d2(ƒ(x))/(d(x)2)=12*x+6

  1. Set the second derivative to zero to find potential inflection points where the concavity might change.

12*x+6=0

  1. Solve for x by isolating the variable.

12*x=−6

x=−1/2

  1. Verify the change in concavity by checking the sign of ƒ(x)″ on either side of x=−1/2 Since ƒ(x)″ is a linear function with a positive slope, it changes from negative to positive at this point.

  2. Calculate the y-coordinate by substituting x=−1/2 back into the original function ƒ(x)

ƒ*(−1/2)=2*(−1/2)3+3*(−1/2)2−180*(−1/2)

ƒ*(−1/2)=2*(−1/8)+3*(1/4)+90

ƒ*(−1/2)=−1/4+3/4+90

ƒ*(−1/2)=1/2+90

ƒ*(−1/2)=90.5

Final Answer

Inflection Point=(−1/2,90.5)


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