Find the Local Maxima and Minima f(x)=3x^4-4x^3

Problem

ƒ(x)=3*x4−4*x3

Solution

1. Find the first derivative of the function to identify critical points.

(d(3)*x4−4*x3)/d(x)=12*x3−12*x2

2. Set the derivative to zero and solve for x to find the critical points.

12*x3−12*x2=0

12*x2*(x−1)=0

x=0,x=1

3. Find the second derivative to apply the Second Derivative Test.

(d(12)*x3−12*x2)/d(x)=36*x2−24*x

4. Evaluate the second derivative at the critical points.

ƒ(0)″=36*(0)2−24*(0)=0

ƒ(1)″=36*(1)2−24*(1)=12

5. Apply the First Derivative Test for x=0 since the Second Derivative Test was inconclusive (ƒ(0)″=0.
   For x<0 (e.g., −1, ƒ′*(−1)=12*(−1)2*(−1−1)=−24<0
   For 0<x<1 (e.g., 0.5, ƒ(0.5)′=12*(0.5)2*(0.5−1)=−1.5<0
   Since the sign of the derivative does not change at x=0 there is no local extremum at x=0

6. Interpret the result for x=1 Since ƒ(1)″=12>0 the function has a local minimum at x=1

7. Calculate the function value at the local minimum.

ƒ(1)=3*(1)4−4*(1)3=−1

Final Answer

Local Minimum: *(1,−1), Local Maxima: None

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