Find without using a calculator the absolute extreme values

Find (without using a calculator) the absolute extreme values of the function on the given interval.

f(x) = x³ 12x² + 21x + 5 on [1, 2]

Solution

given f(x) = x³ 12x² + 21x + 5

differentiate with respect to x.

f \'(x) = 3x^3-1 12*2x^2-1 + 21*1 + 0

f \'(x) = 3x² 24x + 21

f \'(x) = 3(x² 8x +7)

f \'(x) = 3(x 1)(x 7)

for crtical numbers , f \'(x)=0

=>3(x 1)(x 7)=0

=>x=1, x=7

only x=1 lies ininterval  [1, 2]

f(-1) = (-1)³ 12(-1)² + 21(-1) + 5 =-29

f(1) = (1)³ 12(1)² + 21(1) + 5 =15

f(2) = (2)³ 12(2)² + 21(2) + 5 =7

absolute minimum is -29

absolute maximum is 15