If x in R and 1 leq frac{3x^2-7x+8}{x^2+1} le | vedclass

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(D) Given,$1 \leq \frac{3x^2-7x+8}{x^2+1} \leq 2$. Since $x^2+1 > 0$ for all $x \in R$,we can multiply throughout by $(x^2+1)$.First,consider $1 \leq

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$1, 6$

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(D) Given,$1 \leq \frac{3x^2-7x+8}{x^2+1} \leq 2$. Since $x^2+1 > 0$ for all $x \in R$,we can multiply throughout by $(x^2+1)$.First,consider $1 \leq \frac{3x^2-7x+8}{x^2+1} \implies x^2+1 \leq 3x^2-7x+8 \implies 2x^2-7x+7 \geq 0$.The discriminant of $2x^2-7x+7$ is $D = (-7)^2 - 4(2)(7) = 49 - 56 = -7  0$ for all $x \in R$.Next,consider $\frac{3x^2-7x+8}{x^2+1} \leq 2 \implies 3x^2-7x+8 \leq 2x^2+2 \implies x^2-7x+6 \leq 0$.Factoring the quadratic,we get $(x-1)(x-6) \leq 0$.This inequality holds for $x \in [1, 6]$.Thus,the minimum value is $1$ and the maximum value is $6$.

If $x \in R$ and $1 \leq \frac{3x^2-7x+8}{x^2+1} \leq 2$,then the minimum and maximum values of $x$ are respectively.

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