If the function f(x) = frac{t + 3x - x^2}{x - | vedclass

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(C) Given the function $f(x) = \frac{t + 3x - x^2}{x - 4}$.To find the critical points,we calculate the derivative $f'(x)$ using the quotient rule:$f'

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$(-\infty, 4)$

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(C) Given the function $f(x) = \frac{t + 3x - x^2}{x - 4}$.To find the critical points,we calculate the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(x - 4)(3 - 2x) - (t + 3x - x^2)(1)}{(x - 4)^2}$$f'(x) = \frac{3x - 2x^2 - 12 + 8x - t - 3x + x^2}{(x - 4)^2}$$f'(x) = \frac{-x^2 + 8x - (12 + t)}{(x - 4)^2}$For the function to have a local maximum and a local minimum,the derivative $f'(x) = 0$ must have two distinct real roots,and these roots must not be equal to the vertical asymptote $x = 4$.Setting the numerator to zero: $-x^2 + 8x - (12 + t) = 0$,or $x^2 - 8x + (12 + t) = 0$.For two distinct real roots,the discriminant $D$ must be greater than $0$:$D = (-8)^2 - 4(1)(12 + t) > 0$$64 - 48 - 4t > 0$$16 - 4t > 0$$4 > t$,which means $t Also,the root $x = 4$ must not be a solution to the numerator equation:$(4)^2 - 8(4) + 12 + t 
eq 0$$16 - 32 + 12 + t 
eq 0$$-4 + t 
eq 0 \implies t 
eq 4$.Thus,the range of values for $t$ is $t < 4$,which is $(-\infty, 4)$.

If the function $f(x) = \frac{t + 3x - x^2}{x - 4}$,where $t$ is a parameter,has a local maximum and a local minimum,then the range of values of $t$ is:

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