Find the absolute maximum and absolute minimum values of f on the given interval. f(x)=2x3−3x2−12x+1,[−2,3]

Question

Find the absolute maximum and absolute minimum values of f on the given interval.  f(x)=2x3−3x2−12x+1,[−2,3]

Rosemary McBride · Accepted Answer

Step 1  The function is given by  f(x)=2x3−3x2−12x+1,[−2,3]  To evaluate : The absolute maximum and minimum of the function on the given interval.  Step 2  Let us evaluate the critical numbers,  dfdx=0  ⇒ddx(2x3−3x2−12x+1)=0  ⇒6x2−6x−12=0  ⇒x2−x−2=0  ⇒x2−2x+x−2=0  ⇒x(x−2)+1(x−2)=0  ⇒(x+1)(x−2)=0  ⇒x=−1,2 These are critical numbers  Step 3  Now, let us evaluate the values of the function at the critical numbers and at the endpoints of the closed interval [-2,3]  At x=-1,  f(−1)=2×(−1)3−3×(−1)2−12×(−1)+1=8  Absolute maximum   At x=2,  f(2)=2×(2)3−3×(2)2−12×(2)+1=−19  Absolute minimum  At x=-2,  f(−2)=2×(−2)3−3×(−2)2−12×(−2)+1=−3  At x=3,  f(3)=2×(3)3−3×(3)2−12×(3)+1=−8  Step 4  Hence, the absolute maximum and absolute minimum values of the function are :  Absolute maximum value of the function = 8  Absolute minimum value of the function =-19

Find the absolute maximum and absolute minimum values of f

Answered question

Answer & Explanation

New Questions in Algebra I