Find Where Increasing/Decreasing Using Derivatives f(x)=2x^3+3x^2-72x

Problem

ƒ(x)=2*x3+3*x2−72*x

Solution

1. Find the derivative of the function to determine the slope of the tangent line at any point x

d(ƒ(x))/d(x)=6*x2+6*x−72

2. Set the derivative to zero to find the critical points where the function might change direction.

6*x2+6*x−72=0

3. Factor the quadratic equation to solve for the critical values of x

6*(x2+x−12)=0

6*(x+4)*(x−3)=0

x=−4,x=3

4. Test the intervals created by the critical points (−∞,−4) (−4,3) and (3,∞) by plugging a test value into ƒ(x)′
   For x=−5 ƒ′*(−5)=6*(−5+4)*(−5−3)=48>0 (Increasing)
   For x=0 ƒ(0)′=6*(0+4)*(0−3)=−72<0 (Decreasing)
   For x=4 ƒ(4)′=6*(4+4)*(4−3)=48>0 (Increasing)

5. Identify the intervals based on the sign of the derivative.
   The function is increasing where ƒ(x)′>0 and decreasing where ƒ(x)′<0

Final Answer

Increasing: *(−∞,−4)∪(3,∞), Decreasing: *(−4,3)

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