Sponsor Area

Application Of Derivatives

Question
CBSEENMA12035398

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight x minus 1 right parenthesis squared plus 3 comma space space space space straight x space element of space space open square brackets negative 3 comma space space 1 close square brackets



Solution

Here f (x) = (x – 1 )2 + 3
The given function is differentiable for all x in [– 3, 1],
f '(x) = 2 (x – 1)
Now f ' (x) = 0 ⇒ 2 (x – 1) = 0 ⇒ x = [∊ [– 3,1]
f (– 3) = (– 3 – 1)2 + 3 = 19, f (1) = (1 – 1 )2 + 3 = 3
∴  absolute maximum value = 19 and absolute minimum value = 3.