Find the absolute maximum value and the absol | vedclass

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(A) The given function is $f(x) = (x - 1)^{2} + 3$ defined on the interval $[-3, 1]$.First,we find the derivative of the function:$f'(x) = 2(x - 1)$.T

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Absolute maximum value is $19$ and absolute minimum value is $3$.

Vedclass · Answer

(A) The given function is $f(x) = (x - 1)^{2} + 3$ defined on the interval $[-3, 1]$.First,we find the derivative of the function:$f'(x) = 2(x - 1)$.To find the critical points,we set $f'(x) = 0$:$2(x - 1) = 0 \Rightarrow x = 1$.Now,we evaluate the function $f(x)$ at the critical point $x = 1$ and at the endpoints of the interval $x = -3$ and $x = 1$:$f(1) = (1 - 1)^{2} + 3 = 0 + 3 = 3$.$f(-3) = (-3 - 1)^{2} + 3 = (-4)^{2} + 3 = 16 + 3 = 19$.Comparing these values,the absolute maximum value is $19$ at $x = -3$ and the absolute minimum value is $3$ at $x = 1$.

Find the absolute maximum value and the absolute minimum value of the function given by $f(x) = (x - 1)^{2} + 3, x \in [-3, 1]$.

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