The function f(x) = x^3 - 4x^2 + 4x + 3 defin | vedclass

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(A) Given,$f(x) = x^3 - 4x^2 + 4x + 3$ on the interval $[-1, 3]$.First,find the critical points by setting $f'(x) = 0$.$f'(x) = 3x^2 - 8x + 4 = 0$.$(3

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minimum value $-6$ at $x = -1$

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(A) Given,$f(x) = x^3 - 4x^2 + 4x + 3$ on the interval $[-1, 3]$.First,find the critical points by setting $f'(x) = 0$.$f'(x) = 3x^2 - 8x + 4 = 0$.$(3x - 2)(x - 2) = 0$,which gives $x = \frac{2}{3}$ and $x = 2$.Both points lie within the interval $[-1, 3]$.Now,evaluate $f(x)$ at the critical points and the endpoints:$f(-1) = (-1)^3 - 4(-1)^2 + 4(-1) + 3 = -1 - 4 - 4 + 3 = -6$.$f(\frac{2}{3}) = (\frac{2}{3})^3 - 4(\frac{2}{3})^2 + 4(\frac{2}{3}) + 3 = \frac{8}{27} - \frac{16}{9} + \frac{8}{3} + 3 = \frac{8 - 48 + 72 + 81}{27} = \frac{113}{27} \approx 4.18$.$f(2) = (2)^3 - 4(2)^2 + 4(2) + 3 = 8 - 16 + 8 + 3 = 3$.$f(3) = (3)^3 - 4(3)^2 + 4(3) + 3 = 27 - 36 + 12 + 3 = 6$.Comparing these values: $\{-6, 4.18, 3, 6\}$,the minimum value is $-6$ at $x = -1$.

The function $f(x) = x^3 - 4x^2 + 4x + 3$ defined on $[-1, 3]$ has

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