The sum of the maximum and the minimum values | vedclass

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(B) Given the function $f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 4$.First,find the derivative $f'(x)$:$f'(x) = 12x^3 - 6x^2 - 12x + 6$Factorize the derivative

Vedclass · Accepted Answer

$\frac{167}{16}$

Vedclass · Answer

(B) Given the function $f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 4$.First,find the derivative $f'(x)$:$f'(x) = 12x^3 - 6x^2 - 12x + 6$Factorize the derivative:$f'(x) = 6(2x^3 - x^2 - 2x + 1) = 6[x^2(2x - 1) - 1(2x - 1)] = 6(x^2 - 1)(2x - 1) = 6(x - 1)(x + 1)(2x - 1)$.Set $f'(x) = 0$ to find critical points:$6(x - 1)(x + 1)(2x - 1) = 0$.The critical points are $x = 1, x = -1, x = 1/2$.Since we are considering the interval $(0, 2)$,we only consider $x = 1$ and $x = 1/2$ (as $-1$ is outside the interval).Evaluate $f(x)$ at these points:$f(1) = 3(1)^4 - 2(1)^3 - 6(1)^2 + 6(1) + 4 = 3 - 2 - 6 + 6 + 4 = 5$.$f(1/2) = 3(1/16) - 2(1/8) - 6(1/4) + 6(1/2) + 4 = 3/16 - 4/16 - 24/16 + 48/16 + 64/16 = 87/16$.Comparing the values,the maximum value is $87/16$ and the minimum value is $5$.The sum of the maximum and minimum values is $87/16 + 5 = (87 + 80) / 16 = 167/16$.

The sum of the maximum and the minimum values of $f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 4$ in the interval $(0, 2)$ is:

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