Let f(x)=frac{6 x^2-18 x+21}{6 x^2-18 x+17}. | vedclass

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(B) Given, $f(x)=\frac{6 x^2-18 x+21}{6 x^2-18 x+17}$.We can rewrite the function as $f(x)=1+\frac{4}{6 x^2-18 x+17}$.Let $g(x)=6 x^2-18 x+17$. To fin

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$23$

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(B) Given, $f(x)=\frac{6 x^2-18 x+21}{6 x^2-18 x+17}$.We can rewrite the function as $f(x)=1+\frac{4}{6 x^2-18 x+17}$.Let $g(x)=6 x^2-18 x+17$. To find the maximum value of $f(x)$, we need to find the minimum value of $g(x)$.$g'(x)=12 x-18$. Setting $g'(x)=0$, we get $x=\frac{3}{2}$.The minimum value of $g(x)$ is $g(\frac{3}{2}) = 6(\frac{9}{4}) - 18(\frac{3}{2}) + 17 = \frac{27}{2} - 27 + 17 = \frac{27-54+34}{2} = \frac{7}{2}$.Thus, the maximum value of $f(x)$ is $m = 1 + \frac{4}{7/2} = 1 + \frac{8}{7} = \frac{15}{7}$.As $x 	o \pm \infty$, $f(x) 	o 1$. Since $g(x) > 0$ for all $x$, $f(x) > 1$ for all $x \in R$. Thus, $n = 1$.Finally, $14 m - 7 n = 14(\frac{15}{7}) - 7(1) = 30 - 7 = 23$.

Let $f(x)=\frac{6 x^2-18 x+21}{6 x^2-18 x+17}$. If $m$ is the maximum value of $f(x)$ and $f(x) > n$ for all $x \in R$, then $14 m-7 n =$

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