Find the Absolute Max and Min over the Interval f(x)=4x^3-34x^2+60x , 0<x<2.5

Problem

ƒ(x)=4*x3−34*x2+60*x,0≤x≤2.5

Solution

1. Find the derivative of the function to locate critical points.

d(ƒ(x))/d(x)=12*x2−68*x+60

2. Set the derivative to zero to find the critical values of x

12*x2−68*x+60=0

3. Simplify the quadratic equation by dividing all terms by 4.

3*x2−17*x+15=0

4. Apply the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a) to solve for x

x=(17±√(,(−17)2−4*(3)*(15)))/(2*(3))

x=(17±√(,289−180))/6

x=(17±√(,109))/6

5. Calculate the numerical values of the critical points.

(x_1)=(17+√(,109))/6≈4.57

(x_2)=(17−√(,109))/6≈1.09

6. Identify critical points within the interval [0,2.5] Only x≈1.09 is in the interval.

x≈1.093

7. Evaluate the function at the endpoints of the interval.

ƒ(0)=4*(0)3−34*(0)2+60*(0)=0

ƒ(2.5)=4*(2.5)3−34*(2.5)2+60*(2.5)=0

8. Evaluate the function at the critical point x=(17−√(,109))/6

ƒ(1.093)≈4*(1.093)3−34*(1.093)2+60*(1.093)≈30.14

9. Compare the values to determine the absolute maximum and minimum. The values are 0 30.14 and 0

Final Answer

Absolute Max≈30.14,Absolute Min=0

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