Find the Critical Points y=4x^3-48x-8

Problem

y=4*x3−48*x−8

Solution

1. Find the derivative of the function with respect to x using the power rule.

d(y)/d(x)=12*x2−48

2. Set the derivative equal to zero to find the values of x where the slope of the tangent line is zero.

12*x2−48=0

3. Factor out the constant to simplify the quadratic equation.

12*(x2−4)=0

4. Solve for x by factoring the difference of squares or by isolating x2

x2−4=0

x2=4

x=±2

5. Evaluate the original function at these x values to find the corresponding y coordinates.

y(2)=4*(2)3−48*(2)−8=32−96−8=−72

y*(−2)=4*(−2)3−48*(−2)−8=−32+96−8=56

Final Answer

(2,−72),(−2,56)

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