The minimum value of the function f(x) = frac | vedclass

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(D) Let $g(x) = 3x^4 + 8x^3 - 18x^2 + 60$. The function $f(x) = \frac{40}{g(x)}$ is minimum when $g(x)$ is maximum,and $f(x)$ is maximum when $g(x)$ i

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None of these

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(D) Let $g(x) = 3x^4 + 8x^3 - 18x^2 + 60$. The function $f(x) = \frac{40}{g(x)}$ is minimum when $g(x)$ is maximum,and $f(x)$ is maximum when $g(x)$ is minimum.To find the extrema of $g(x)$,we find its derivative: $g'(x) = 12x^3 + 24x^2 - 36x$.Setting $g'(x) = 0$,we get $12x(x^2 + 2x - 3) = 0$,which factors to $12x(x+3)(x-1) = 0$.The critical points are $x = 0, x = 1, x = -3$.We evaluate $g(x)$ at these points:$g(0) = 60$.$g(1) = 3(1)^4 + 8(1)^3 - 18(1)^2 + 60 = 3 + 8 - 18 + 60 = 53$.$g(-3) = 3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60 = 3(81) + 8(-27) - 18(9) + 60 = 243 - 216 - 162 + 60 = -75$.Since $g(x) 	o \infty$ as $x 	o \pm \infty$,the global minimum of $g(x)$ is $-75$ at $x = -3$. However,$f(x)$ is defined as $40/g(x)$.Checking the values of $f(x)$ at critical points:$f(0) = 40/60 = 2/3$.$f(1) = 40/53$.$f(-3) = 40/(-75) = -8/15$.As $x 	o \pm \infty$,$f(x) 	o 0$.The minimum value of the function is $-8/15$. Since this is not among the options,the correct choice is $D$.

The minimum value of the function $f(x) = \frac{40}{3x^4 + 8x^3 - 18x^2 + 60}$ is:

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